4_The Putnam Mathematica Competition (2001-2008).AMS

The 62nd William Lowell Putnam Mathematical CompetitionSaturday, December 1, 2001

A-1 Consider a set

and a binary operation , i.e., for each , . Assume for all . Prove that for all .

A-2 You have coins . For each , ! is bi-ased so that, when tossed, it has probability "#$&%' ()" of fallings heads. If the * coins are tossed, what is theprobability that the number of heads is odd? Expressthe answers as a rational function of * .

A-3 For each integer + , consider the polynomial,.- &/ 0 /$12)&%'+3(54 / (6+725% For what values of + is

,8- &/ the product of two non-constant polynomials with integer coefficients?

A-4 Triangle 9:; has an area 1. Points < => lie, respec-tively, on sides :; , ?9 , 9?: such that 9?< bisects : =at point @ , : = bisects > at point

, and > bisects9< at point A . Find the area of the triangle @ A .

A-5 Prove that there are unique positive integers , * suchthat B8 2) (C" %ED'DF" .

A-6 Can an arc of a parabola inside a circle of radius 1 havea length greater than 4?

B-1 Let * be an even positive integer. Write the numbers" % G * in the squares of an *IHJ* grid so that the -th row, from left to right, is

6 K2L" &* ()" 6 ;2M" *N(M% O 6 ;2M" *P(5* Color the squares of the grid so that half of the squaresin each row and in each column are red and the otherhalf are black (a checkerboard coloring is one possi-bility). Prove that for each coloring, the sum of thenumbers on the red squares is equal to the sum of thenumbers on the black squares.

B-2 Find all pairs of real numbers 6/ QE satisfying the sys-tem of equations

"/ (

"% Q 6/ (SR Q &RE/ ( Q

"/ 2

"% Q %$ Q 1 2T/ 1 U

B-3 For any positive integer * , let VW*YX denote the closest in-teger to Z * . Evaluate

[\^]8

%E_ ` (M%aY_ ^`%

B-4 Let

denote the set of rational numbers different fromb 2;" D "dc . Define e7f hgiby e.6/ P /j23"#'/ .

Prove or disprove that

[k]8 eml

^n o)p$

where e l n denotes e composed with itself * times.

B-5 Let and be real numbers in the interval 6D "#E% ,and let q be a continuous real-valued function suchthat qYrqs6/ Otu qs6/ ( / for all real / . Prove thatqs6/ Cv / for some constant v .

B-6 Assume that $w. is an increasing sequence of pos-itive real numbers such that xryrz F#'* D . Mustthere exist infinitely many positive integers * such that a ( ^B | % for " % G *K2)" ?

This is trial versionwww.adultpdf.com

The 63rd William Lowell Putnam Mathematical CompetitionSaturday, December 7, 2002

A1 Let k be a fixed positive integer. Then-th derivative of1

xk−1has the form Pn(x)

(xk−1)n+1 wherePn(x) is a polyno-mial. FindPn(1).

A2 Given any five points on a sphere, show that some fourof them must lie on a closed hemisphere.

A3 Let n ≥ 2 be an integer andTn be the number of non-empty subsetsS of 1, 2, 3, . . . , n with the propertythat the average of the elements ofS is an integer. ProvethatTn − n is always even.

A4 In Determinant Tic-Tac-Toe, Player 1 enters a 1 in anempty3× 3 matrix. Player 0 counters with a 0 in a va-cant position, and play continues in turn until the3× 3matrix is completed with five 1’s and four 0’s. Player0 wins if the determinant is 0 and player 1 wins other-wise. Assuming both players pursue optimal strategies,who will win and how?

A5 Define a sequence bya0 = 1, together with the rulesa2n+1 = an anda2n+2 = an + an+1 for each inte-ger n ≥ 0. Prove that every positive rational numberappears in the set

an−1

an: n ≥ 1

=

11,12,21,13,32, . . .

.

A6 Fix an integerb ≥ 2. Let f(1) = 1, f(2) = 2, andfor eachn ≥ 3, definef(n) = nf(d), whered is thenumber of base-b digits ofn. For which values ofb does

∞∑n=1

1f(n)

converge?

B1 Shanille O’Keal shoots free throws on a basketballcourt. She hits the first and misses the second, andthereafter the probability that she hits the next shot isequal to the proportion of shots she has hit so far. Whatis the probability she hits exactly 50 of her first 100shots?

B2 Consider a polyhedron with at least five faces such thatexactly three edges emerge from each of its vertices.Two players play the following game:

Each player, in turn, signs his or hername on a previously unsigned face. Thewinner is the player who first succeeds insigning three faces that share a commonvertex.

Show that the player who signs first will always win byplaying as well as possible.

B3 Show that, for all integersn > 1,

12ne

<1e−

(1− 1

n

)n

<1ne

.

B4 An integern, unknown to you, has been randomlychosen in the interval[1, 2002] with uniform probabil-ity. Your objective is to selectn in an odd number ofguesses. After each incorrect guess, you are informedwhethern is higher or lower, and youmust guess aninteger on your next turn among the numbers that arestill feasibly correct. Show that you have a strategy sothat the chance of winning is greater than2/3.

B5 A palindrome in baseb is a positive integer whose base-b digits read the same backwards and forwards; for ex-ample,2002 is a 4-digit palindrome in base 10. Notethat 200 is not a palindrome in base 10, but it is the 3-digit palindrome 242 in base 9, and 404 in base 7. Provethat there is an integer which is a 3-digit palindrome inbaseb for at least 2002 different values ofb.

B6 Letp be a prime number. Prove that the determinant ofthe matrix x y z

xp yp zp

xp2yp2

zp2

is congruent modulop to a product of polynomials ofthe formax + by + cz, wherea, b, c are integers. (Wesay two integer polynomials are congruent modulop ifcorresponding coefficients are congruent modulop.)


The 64th William Lowell Putnam Mathematical CompetitionSaturday, December 6, 2003

A1 Let be a fixed positive integer. How many ways arethere to write as a sum of positive integers, , with an arbitrary positive integerand ? For example, with there are four ways: 4, 2+2, 1+1+2, 1+1+1+1.

A2 Let !!!"$# and % % &!!! % # be nonnegativereal numbers. Show that' " #( *)"# ' % % % #( *)"#,+ ' % -( ' % ( ' # % #(/. )"# !

A3 Find the minimum value of02135476 98: 16 ;"< 4=6 98:&; 6 1> 8 6 8 1 8 6?0for real numbers

6.

A4 Suppose that @ % AB"CD"EF"G are real numbers, IHKJand CLHMJ , such that0 6 % 6 9A 0 0 C 6 9E 6 NG 0for all real numbers

6. Show that0 % =O PA 0 0 E 7O PCQG 0 !

A5 A Dyck -path is a lattice path of upsteps' ( and

downsteps' O ( that starts at the origin R and never

dips below the6

-axis. A return is a maximal sequenceof contiguous downsteps that terminates on the

6-axis.

For example, the Dyck 5-path illustrated has two re-turns, of length 3 and 1 respectively.

O

Show that there is a one-to-one correspondence be-tween the Dyck -paths with no return of even lengthand the Dyck

' O ( -paths.

A6 For a set S of nonnegative integers, let TBU ' ( denotethe number of ordered pairs

'WV B V ( such thatV YX S ,V ZX S ,

V [H V , andV \ V ]^ . Is it possible to

partition the nonnegative integers into two sets C andE in such a way that T_ ' ( T` ' ( for all ?

B1 Do there exist polynomials 'a6 ( % 'b6 ( "A 'bc ( "d 'ac ( suchthat e 6fc 6 c M 'a6 ( A 'bc ( % 'b6 ( d 'bc (holds identically?

B2 Let be a positive integer. Starting with the sequence g !!! # , form a new sequence of O entriesgh 7i2 !!! "#jk"#l5#jm2n by taking the averages of two con-secutive entries in the first sequence. Repeat the aver-aging of neighbors on the second sequence to obtain athird sequence of Opo

entries, and continue until thefinal sequence produced consists of a single number

6 # .Show that

6 #]q o$r .

B3 Show that for each positive integer n,

ts$ #uvxw y 8z|$ o !!!~ rP !(Here y 8-z denotes the least common multiple, and ~ 6 denotes the greatest integer 6

.)

B4 Let 'a ( M h % g A 9d M 'W O T ( 'a O T ( 'W O T g ( 'a O T h (where @ % "ABd are integers, KHJ . Show that ifT Q T is a rational number and T \ T pH T g T h ,then T T is a rational number.

B5 Let CDE , and G be equidistant points on the circumfer-ence of a circle of unit radius centered at R , and let be any point in the circle’s interior. Let @ % "A be the dis-tance from to CE"G , respectively. Show that thereis a triangle with side lengths @ % A , and that the area ofthis triangle depends only on the distance from to R .

B6 Let

'a6 ( be a continuous real-valued function definedon the interval + J . . Show that

0 'b6 ( 'ac ( 0 d 6 d c

0 'b6 ( 0 d 6 !



A1 Basketball star Shanille O’Keal’s team statisticiankeeps track of the number,S(N), of successful freethrows she has made in her firstN attempts of the sea-son. Early in the season,S(N) was less than 80% ofN , but by the end of the season,S(N) was more than80% ofN . Was there necessarily a moment in betweenwhenS(N) was exactly 80% ofN?

A2 For i = 1, 2 let Ti be a triangle with side lengthsai, bi, ci, and areaAi. Suppose thata1 ≤ a2, b1 ≤b2, c1 ≤ c2, and thatT2 is an acute triangle. Does itfollow thatA1 ≤ A2?

A3 Define a sequenceun∞n=0 by u0 = u1 = u2 = 1, andthereafter by the condition that

det(

un un+1

un+2 un+3

)= n!

for all n ≥ 0. Show thatun is an integer for alln. (Byconvention,0! = 1.)

A4 Show that for any positive integern, there is an integerN such that the productx1x2 · · ·xn can be expressedidentically in the form

x1x2 · · ·xn =N∑

i=1

ci(ai1x1 + ai2x2 + · · ·+ ainxn)n

where theci are rational numbers and eachaij is one ofthe numbers−1, 0, 1.

A5 An m × n checkerboard is colored randomly: eachsquare is independently assigned red or black withprobability1/2. We say that two squares,p andq, are inthe same connected monochromatic component if thereis a sequence of squares, all of the same color, startingatp and ending atq, in which successive squares in thesequence share a common side. Show that the expectednumber of connected monochromatic regions is greaterthanmn/8.

A6 Suppose thatf(x, y) is a continuous real-valued func-tion on the unit square0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Showthat∫ 1

0

(∫ 1

0

f(x, y)dx

)2

dy +∫ 1

0

(∫ 1

0

f(x, y)dy

)2

dx

≤(∫ 1

0

∫ 1

0

f(x, y)dx dy

)2

+∫ 1

0

∫ 1

0

(f(x, y))2 dx dy.

B1 Let P (x) = cnxn + cn−1xn−1 + · · · + c0 be a poly-

nomial with integer coefficients. Suppose thatr is arational number such thatP (r) = 0. Show that thennumbers

cnr, cnr2 + cn−1r, cnr3 + cn−1r2 + cn−2r,

. . . , cnrn + cn−1rn−1 + · · ·+ c1r

are integers.

B2 Letm andn be positive integers. Show that

(m + n)!(m + n)m+n

<m!mm

n!nn

.

B3 Determine all real numbersa > 0 for which there ex-ists a nonnegative continuous functionf(x) defined on[0, a] with the property that the region

R = (x, y); 0 ≤ x ≤ a, 0 ≤ y ≤ f(x)

has perimeterk units and areak square units for somereal numberk.

B4 Let n be a positive integer,n ≥ 2, and putθ = 2π/n.Define pointsPk = (k, 0) in the xy-plane, fork =1, 2, . . . , n. Let Rk be the map that rotates the planecounterclockwise by the angleθ about the pointPk. LetR denote the map obtained by applying, in order,R1,then R2, . . . , thenRn. For an arbitrary point(x, y),find, and simplify, the coordinates ofR(x, y).

B5 Evaluate

limx→1−

∞∏n=0

(1 + xn+1

1 + xn

)xn

.

B6 LetA be a non-empty set of positive integers, and letN(x) denote the number of elements ofA not exceed-ing x. Let B denote the set of positive integersb thatcan be written in the formb = a − a′ with a ∈ A anda′ ∈ A. Let b1 < b2 < · · · be the members ofB, listedin increasing order. Show that if the sequencebi+1 − bi

is unbounded, then

limx→∞

N(x)/x = 0.



A1 Show that every positive integer is a sum of one or morenumbers of the form2r3s, wherer ands are nonneg-ative integers and no summand divides another. (Forexample, 23 = 9 + 8 + 6.)

A2 Let S = (a, b)|a = 1, 2, . . . , n, b = 1, 2, 3. A rooktour of S is a polygonal path made up of line segmentsconnecting pointsp1, p2, . . . , p3n in sequence such that

(i) pi ∈ S,

(ii) pi andpi+1 are a unit distance apart, for1 ≤ i <3n,

(iii) for eachp ∈ S there is a uniquei such thatpi = p.How many rook tours are there that begin at(1, 1)and end at(n, 1)?

(An example of such a rook tour forn = 5 was depictedin the original.)

A3 Let p(z) be a polynomial of degreen, all of whose ze-ros have absolute value 1 in the complex plane. Putg(z) = p(z)/zn/2. Show that all zeros ofg′(z) = 0have absolute value 1.

A4 Let H be ann × n matrix all of whose entries are±1and whose rows are mutually orthogonal. SupposeHhas ana × b submatrix whose entries are all1. Showthatab ≤ n.

A5 Evaluate ∫ 1

0

ln(x + 1)x2 + 1

dx.

A6 Let n be given,n ≥ 4, and suppose thatP1, P2, . . . , Pn

aren randomly, independently and uniformly, chosenpoints on a circle. Consider the convexn-gon whosevertices arePi. What is the probability that at least oneof the vertex angles of this polygon is acute?

B1 Find a nonzero polynomialP (x, y) such thatP (bac, b2ac) = 0 for all real numbersa. (Note:bνc is the greatest integer less than or equal toν.)

B2 Find all positive integersn, k1, . . . , kn such thatk1 +· · ·+ kn = 5n− 4 and

1k1

+ · · ·+ 1kn

= 1.

B3 Find all differentiable functionsf : (0,∞) → (0,∞)for which there is a positive real numbera such that

f ′(a

x

)=

x

f(x)

for all x > 0.B4 For positive integersm andn, let f(m,n) denote the

number ofn-tuples (x1, x2, . . . , xn) of integers suchthat|x1|+ |x2|+ · · ·+ |xn| ≤ m. Show thatf(m,n) =f(n, m).

B5 LetP (x1, . . . , xn) denote a polynomial with real coef-ficients in the variablesx1, . . . , xn, and suppose that(

∂2

∂x21

+ · · ·+ ∂2

∂x2n

)P (x1, . . . , xn) = 0 (identically)

and that

x21 + · · ·+ x2

n dividesP (x1, . . . , xn).

Show thatP = 0 identically.

B6 LetSn denote the set of all permutations of the numbers1, 2, . . . , n. For π ∈ Sn, let σ(π) = 1 if π is an evenpermutation andσ(π) = −1 if π is an odd permutation.Also, let ν(π) denote the number of fixed points ofπ.Show that∑

π∈Sn

σ(π)ν(π) + 1

= (−1)n+1 n

n + 1.



A1 Find the volume of the region of points (x, y, z) suchthat

(x2 + y2 + z2 + 8)2 ≤ 36(x2 + y2).

A2 Alice and Bob play a game in which they take turnsremoving stones from a heap that initially has n stones.The number of stones removed at each turn must be oneless than a prime number. The winner is the player whotakes the last stone. Alice plays first. Prove that thereare infinitely many n such that Bob has a winning strat-egy. (For example, if n = 17, then Alice might take6 leaving 11; then Bob might take 1 leaving 10; thenAlice can take the remaining stones to win.)

A3 Let 1, 2, 3, . . . , 2005, 2006, 2007, 2009, 2012, 2016, . . .be a sequence defined by xk = k for k = 1, 2, . . . , 2006and xk+1 = xk + xk−2005 for k ≥ 2006. Show that thesequence has 2005 consecutive terms each divisible by2006.

A4 Let S = 1, 2, . . . , n for some integer n > 1. Say apermutation π of S has a local maximum at k ∈ S if

(i) π(k) > π(k + 1) for k = 1;

(ii) π(k − 1) < π(k) and π(k) > π(k + 1) for 1 <k < n;

(iii) π(k − 1) < π(k) for k = n.

(For example, if n = 5 and π takes values at 1, 2, 3, 4, 5of 2, 1, 4, 5, 3, then π has a local maximum of 2 at k =1, and a local maximum of 5 at k = 4.) What is theaverage number of local maxima of a permutation of S,averaging over all permutations of S?

A5 Let n be a positive odd integer and let θ be a real numbersuch that θ/π is irrational. Set ak = tan(θ + kπ/n),k = 1, 2, . . . , n. Prove that

a1 + a2 + · · ·+ an

a1a2 · · · an

is an integer, and determine its value.

A6 Four points are chosen uniformly and independently atrandom in the interior of a given circle. Find the proba-bility that they are the vertices of a convex quadrilateral.

B1 Show that the curve x3 + 3xy + y3 = 1 contains onlyone set of three distinct points, A, B, and C, which arevertices of an equilateral triangle, and find its area.

B2 Prove that, for every set X = x1, x2, . . . , xn of nreal numbers, there exists a non-empty subset S of Xand an integer m such that∣∣∣∣∣m +

∑s∈S

s

∣∣∣∣∣ ≤ 1n + 1

.

B3 Let S be a finite set of points in the plane. A linearpartition of S is an unordered pair A,B of subsets ofS such that A ∪ B = S, A ∩ B = ∅, and A and Blie on opposite sides of some straight line disjoint fromS (A or B may be empty). Let LS be the number oflinear partitions of S. For each positive integer n, findthe maximum of LS over all sets S of n points.

B4 Let Z denote the set of points in Rn whose coordinatesare 0 or 1. (Thus Z has 2n elements, which are thevertices of a unit hypercube in Rn.) Given a vector sub-space V of Rn, let Z(V ) denote the number of membersof Z that lie in V . Let k be given, 0 ≤ k ≤ n. Findthe maximum, over all vector subspaces V ⊆ Rn ofdimension k, of the number of points in V ∩ Z.) [Ed-itorial note: the proposers probably intended to writeZ(V ) for “the number of points in V ∩ Z”, but thischanges nothing.]

B5 For each continuous function f : [0, 1] → R, let I(f) =∫ 1

0x2f(x) dx and J(x) =

∫ 1

0x (f(x))2 dx. Find the

maximum value of I(f)− J(f) over all such functionsf .

B6 Let k be an integer greater than 1. Suppose a0 > 0, anddefine

an+1 = an +1

k√

an

for n > 0. Evaluate

limn→∞

ak+1n

nk.



A1 Find all values ofα for which the curvesy = αx2 +αx + 1

24andx = αy2 + αy + 1

24are tangent to each

other.

A2 Find the least possible area of a convex set in the planethat intersects both branches of the hyperbolaxy = 1and both branches of the hyperbolaxy = −1. (A setSin the plane is calledconvex if for any two points inS

the line segment connecting them is contained inS.)

A3 Let k be a positive integer. Suppose that the integers1, 2, 3, . . . , 3k + 1 are written down in random order.What is the probability that at no time during this pro-cess, the sum of the integers that have been written upto that time is a positive integer divisible by 3? Youranswer should be in closed form, but may include fac-torials.

A4 A repunit is a positive integer whose digits in base 10are all ones. Find all polynomialsf with real coeffi-cients such that ifn is a repunit, then so isf(n).

A5 Suppose that a finite group has exactlyn elements oforderp, wherep is a prime. Prove that eithern = 0 orp dividesn + 1.

A6 A triangulation T of a polygonP is a finite collectionof triangles whose union isP , and such that the inter-section of any two triangles is either empty, or a sharedvertex, or a shared side. Moreover, each side is a sideof exactly one triangle inT . Say thatT is admissibleif every internal vertex is shared by 6 or more triangles.For example, [figure omitted.] Prove that there is anintegerMn, depending only onn, such that any admis-sible triangulation of a polygonP with n sides has atmostMn triangles.

B1 Letf be a polynomial with positive integer coefficients.Prove that ifn is a positive integer, thenf(n) divides

f(f(n) + 1) if and only if n = 1. [Editor’s note: onemust assumef is nonconstant.]

B2 Suppose thatf : [0, 1] → R has a continuous derivativeand that

∫ 1

0f(x) dx = 0. Prove that for everyα ∈

(0, 1),

∣

∣

∣

∣

∫ α

0

f(x) dx

∣

∣

∣

∣

≤ 1

8max

0≤x≤1|f ′(x)|.

B3 Letx0 = 1 and forn ≥ 0, let xn+1 = 3xn + ⌊xn

√5⌋.

In particular,x1 = 5, x2 = 26, x3 = 136, x4 = 712.Find a closed-form expression forx2007. (⌊a⌋ meansthe largest integer≤ a.)

B4 Let n be a positive integer. Find the number of pairsP, Q of polynomials with real coefficients such that

(P (X))2 + (Q(X))2 = X2n + 1

anddeg P > deg Q.

B5 Let k be a positive integer. Prove that there exist poly-nomials P0(n), P1(n), . . . , Pk−1(n) (which may de-pend onk) such that for any integern,

⌊n

k

⌋k

= P0(n) + P1(n)⌊n

k

⌋

+ · · · + Pk−1(n)⌊n

k

⌋k−1

.

(⌊a⌋ means the largest integer≤ a.)

B6 For each positive integern, let f(n) be the number ofways to maken! cents using an unordered collection ofcoins, each worthk! cents for somek, 1 ≤ k ≤ n.Prove that for some constantC, independent ofn,

nn2/2−Cne−n2/4 ≤ f(n) ≤ nn2/2+Cne−n2/4.



A1 Let f : R2 → R be a function such thatf(x, y) +

f(y, z) + f(z, x) = 0 for all real numbersx, y, andz.Prove that there exists a functiong : R → R such thatf(x, y) = g(x) − g(y) for all real numbersx andy.

A2 Alan and Barbara play a game in which they take turnsfilling entries of an initially empty2008 × 2008 array.Alan plays first. At each turn, a player chooses a realnumber and places it in a vacant entry. The game endswhen all the entries are filled. Alan wins if the determi-nant of the resulting matrix is nonzero; Barbara wins ifit is zero. Which player has a winning strategy?

A3 Start with a finite sequencea1, a2, . . . , an of positiveintegers. If possible, choose two indicesj < k suchthat aj does not divideak, and replaceaj andak bygcd(aj , ak) andlcm(aj , ak), respectively. Prove that ifthis process is repeated, it must eventually stop and thefinal sequence does not depend on the choices made.(Note: gcd means greatest common divisor and lcmmeans least common multiple.)

A4 Definef : R → R by

f(x) =

x if x ≤ e

xf(lnx) if x > e.

Does∑

∞

n=11

f(n) converge?

A5 Let n ≥ 3 be an integer. Letf(x) and g(x) bepolynomials with real coefficients such that the points(f(1), g(1)), (f(2), g(2)), . . . , (f(n), g(n)) in R

2 arethe vertices of a regularn-gon in counterclockwise or-der. Prove that at least one off(x) andg(x) has degreegreater than or equal ton − 1.

A6 Prove that there exists a constantc > 0 such that in ev-ery nontrivial finite groupG there exists a sequence oflength at mostc ln |G| with the property that each el-ement ofG equals the product of some subsequence.

(The elements ofG in the sequence are not required tobe distinct. Asubsequence of a sequence is obtainedby selecting some of the terms, not necessarily consec-utive, without reordering them; for example,4, 4, 2 is asubsequence of2, 4, 6, 4, 2, but2, 2, 4 is not.)

B1 What is the maximum number of rational points that canlie on a circle inR2 whose center is not a rational point?(A rational point is a point both of whose coordinatesare rational numbers.)

B2 LetF0(x) = lnx. Forn ≥ 0 andx > 0, letFn+1(x) =∫ x

0Fn(t) dt. Evaluate

limn→∞

n!Fn(1)

lnn.

B3 What is the largest possible radius of a circle containedin a 4-dimensional hypercube of side length 1?

B4 Let p be a prime number. Leth(x) be apolynomial with integer coefficients such thath(0), h(1), . . . , h(p2 − 1) are distinct modulop2.Show that h(0), h(1), . . . , h(p3 − 1) are distinctmodulop3.

B5 Find all continuously differentiable functionsf : R →R such that for every rational numberq, the numberf(q) is rational and has the same denominator asq.(The denominator of a rational numberq is the uniquepositive integerb such thatq = a/b for some integerawith gcd(a, b) = 1.) (Note: gcd means greatest com-mon divisor.)

B6 Letn andk be positive integers. Say that a permutationσ of 1, 2, . . . , n is k-limited if |σ(i) − i| ≤ k for alli. Prove that the number ofk-limited permutations of1, 2, . . . , n is odd if and only ifn ≡ 0 or 1 (mod2k + 1).


Solutions to the 62nd William Lowell Putnam Mathematical CompetitionSaturday, December 1, 2001

Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A–1 The hypothesis implies

for all

(by replacing

by

), and hence for all

!"(using

#$%).

A–2 Let &#' denote the desired probability. Then &( *),+.- ,and, for /0 ) ,& ' 21 3 /3 /54 )6 & '879( 4 1 )3 /4 ):6 )<; & '879( 21 3 / ;=)3 /54 )6 & '879( 4 )3 /54 )?>The recurrence yields &A@ 3 +CB , &#D E-.+.F , and by asimple induction, one then checks that for general / onehas & ' / +: 3 /54 ), .Note: Richard Stanley points out the following nonin-ductive argument. Put G HJI 'KML ( H 4 3.N +: 3.N 4 ), ;then the coefficient of

HPOin G HP is the probability of

getting exactly Q heads. Thus the desired number is G R),S; G T;U),R+ 3 , and both values of G can be com-puted directly: G ),V) , andG T;U),$ )-XW -BXWZY,Y[YCW 3 / ;)3 /54 ) )3 /54 ) >

A–3 By the quadratic formula, if &]\ HP^`_ , thenH @ acb 3Cd 3 a 4 3 , and hence the four roots of & \ are

given byegf b d ahb d 3i . If &#\ factors into two

nonconstant polynomials over the integers, then somesubset of

consisting of one or two elements form the

roots of a polynomial with integer coefficients.First suppose this subset has a single element, sayd ajb d 3 ; this element must be a rational number.Then

d ahb d 3 @ 3 4 aEb 3Cd 3 a is an integer,so a is twice a perfect square, say a 3 / @ . But thend ab d 3 V / b ), d 3 is only rational if / b ) , i.e.,if a 3 .Next, suppose that the subset contains two elements;then we can take it to be one of

f d a*b d 3ki , f d 3 bd a i orf b d a 4 d 3 i . In all cases, the sum and the

product of the elements of the subset must be a ratio-nal number. In the first case, this means 3 d a Vl ,so a is a perfect square. In the second case, we have3 d 3 l , contradiction. In the third case, we have d a 4 d 3 @ el , or a 4 3 4 3Cd 3 a %l , whichmeans that a is twice a perfect square.We conclude that & \ H factors into two nonconstantpolynomials over the integers if and only if a is eithera square or twice a square.

Note: a more sophisticated interpretation of this argu-ment can be given using Galois theory. Namely, if ais neither a square nor twice a square, then the numberfields

lm d a andlm d 3 are distinct quadratic fields,

so their compositum is a number field of degree 4,whose Galois group acts transitively on

f b d a=b d 3i .Thus & \ is irreducible.

A–4 Choose n oCTp so that qUr nMsUr tu eo r uv s p rUs , and let w x yz$ denote the area of trianglex^yz . Then w u sq w u|t q since the tri-angles have the same altitude and base. Alsow u sUq| sq + sUr w u sUr| 2); n , and w~qUr t qr + sUr r t<+ r u| w u sUr n );o[ (e.g., by thelaw of sines). Adding this all up yields)< w u sUq:4=w u s t k4=w~qUr t 3 R)$; n 4n R)$;o, 3 ; n ; n oor n R) 4 o,V) . Similarly

oCR) 4 pp,R) 4n ) .Let Gw _8Z w _: be the function given byG H`)[+:) 4 HP ; then G G G n Rm n . However,G H is strictly decreasing in

H, so G G HP is increas-

ing and G G G HR is decreasing. Thus there is at mostoneH

such that G G G HR H ; in fact, since the equa-tion G C has a positive root

;) 4 d B.+ 3 , wemust have n op .We now compute w u s t u|t<+u r w u sUr ,w u sU! sU + s tU w u s t .+ 3 , analogouslyw~sUr w~r u .+ 3 , and w~ 9 w u sUr ;w u sU! ; w~sUr ; w~r u [*);-.C+ 3 ,E 7 D .

Note: the key relation n ) 4 o[^) can also be de-rived by computing using homogeneous coordinates orvectors.

A–5 Suppose '#( ;2 4 )[ ' 3 _._8) . Notice that '#( 4=w 4 )[ ' ;) is a multiple of

; thus

divides3 _C_ 3 3 W F W )C) W )[- .

Since 3 _C_:) is divisible by 3, we must haveh) -.

, otherwise one of '#( and

4 ), ' is a mul-tiple of 3 and the other is not, so their difference cannotbe divisible by 3. Now

'#( )X -. , so we musthave

4 )[ ' V)X ^-. , which forces / to be even,and in particular at least 2.

If

is even, then '#( ;U 4 ), ' ¡;U 4 ), ' ¢ .

Since / is even,;U 4 ), ' £;U)X ¢ . Since


3 _._8)UE)¤ ¢ , this is impossible. Thus

is odd,and so must divide

),_C_:)m2F W ).) W )[- . Moreover, '#( ;¥ 4 ), ' ¢ , soUV)¤ ¢

.

Of the divisors ofF W )C) W ),- , those congruent to 1 mod

3 are precisely those not divisible by 11 (since 7 and 13are both congruent to 1 mod 3). Thus

divides

F W )[- .Now

)X ¢is only possible if

divides

),-.

We cannot have¤j)

, since); 3 ' ¦ 3 _._8) for any/ . Thus the only possibility is*§)[-

. One eas-ily checks that

=£),-8 / 3 is a solution; all thatremains is to check that no other / works. In fact,if /¨0 3 , then

)[- '#( 3 _C_:))Z ©. . But),- '#( )[- ©C since / is even, contradiction.Thus

5e)[-: / 3 is the unique solution.

Note: once one has that / is even, one can use that3 _._ 3 ª '#( 4 ); 4 ), ' is divisible by 4 )

to rule out cases.

A–6 The answer is yes. Consider the arc of the parabola« ¬u|H @ inside the circleH @ 4 « ;%), @ j) , where

we initially assume thatu 0 )[+ 3 . This intersects the

circle in three points,_:_.

and b d 3 u;=)[+u, 3 u;),R+u|

. We claim that foru

sufficiently large, thelength ® of the parabolic arc between

_:_.and d 3 u;),+u5[ 3 u¤;X),R+u| is greater than 3 , which im-

plies the desired result by symmetry. We express ® us-ing the usual formula for arclength:® ¯ @° 7#(²± °³ ´ ) 4 3 u|H @?µ H )3 u ¯ @ @° 7#(³ ´ ) 4 H @ µ H 3 4 )3 u·¶ ¯ @ @° 79(³ ´ ) 4 H @ ;HP µ H; 3¹¸ where we have artificially introduced

;!Hinto the inte-

grand in the last step. Now, forHº=_

,´ ) 4 H @ ;!H )d ) 4 H @ 4 H 0 )3 d ) 4 H @ º )3 H 4 ),»since ¼¾½³ µ H+: 3 H 4 ),

diverges, so does¼]½³ d ) 4 H @ ;H µ H . Hence, for sufficiently largeu, we have ¼ @ @° 7#(³ d ) 4 H @ ; H µ H 0 3 , and hence®0 3 .

Note: a numerical computation shows that one musttakeu 0 -M¢ > F to obtain ®0 3 , and that the maximum

value of ® is about¢ > _C_ 3 F , achieved for

u%¿%À¹¢ > ) .B–1 Let (resp. s ) denote the set of red (resp. black)

squares in such a coloring, and foro Ás , letG o, /4Â o[ 4 ) denote the number written in square

o,

where_Ã G o, Â o[Ã / ;) . Then it is clear that the

value of G o, depends only on the row ofo, while the

value of Â o, depends only on the column ofo. Since

every row contains exactly / + 3 elements of and / + 3elements of s , Ä

ÅÇÆÉÈ G o[ÄÅÆMÊ G o[ >

Similarly, because every column contains exactly / + 3elements of and / + 3 elements of s ,Ä

ÅÇÆMÈ Â o[ÄÅÆMÊ Â o[ >

It follows thatÄÅÆMÈ G o[ /4ËÂ o, 4 )

ÄÅÆMÊ G o[ /4XÂ o[ 4 ).

as desired.

Note: Richard Stanley points out a theorem of Ryser(see Ryser, Combinatorial Mathematics, Theorem 3.1)that can also be applied. Namely, if

uand s are

_;=)matrices with the same row and column sums, thenthere is a sequence of operations on 3 W 3 matrices ofthe form 1 _h))%_ 6 1 ) __¬) 6or vice versa, which transforms

uinto s . If we iden-

tify 0 and 1 with red and black, then the given color-ing and the checkerboard coloring both satisfy the sumcondition. Since the desired result is clearly true for thecheckerboard coloring, and performing the matrix op-erations does not affect this, the desired result followsin general.

B–2 By adding and subtracting the two given equations, weobtain the equivalent pair of equations3 +.H%H 4 )[_.H @ « @ 4 B « ),+ « %BCH 4 ),_CH @ « @ 4 « >Multiplying the former by

Hand the latter by « , then

adding and subtracting the two resulting equations, weobtain another pair of equations equivalent to the givenones, -|VH 4 « )¡HÌ; « >It follows that

H- (± 4 ),+ 3 and « - (R± ;=),R+ 3is the unique solution satisfying the given equations.

B–3 Since N ;%),+ 3 @ N @ ; N 4 )[+M¢ and

N 4 ),+ 3 @ N @ 4 N 4 )[+M¢ , we have that ÍT/?Î N if and only if

2This is trial versionwww.adultpdf.com

N @ ; N 4 )Ã / Ã N @ 4 N . Hence½Ä' L ( 3CÏ 'Ð 4 3 7 Ï 'Ð3 ' ½ÄKÉL (Ä':Ñ Ï 'Ð LK 3CÏ 'Ð 4 3 7 Ï 'Ð3 ' ½ÄKÉL (KÉÒ KÄ' LK Ò 7 K #( 3 K 4 3 7 K3 ' ½ÄKÉL ( 3 K 4 3 7 K 3 7 K Ò K ; 3 7 K Ò 7 K ½ÄKÉL ( 3 7 KCÓÔK 7 @Õ ; 3 7 KCÓÔK @Õ ½ÄKÉL ( 3 7 K.ÓK 7 @Õ ; ½

ÄKÉL D 3 7 K.ÓÔK 7 @Õ- >Alternate solution: rewrite the sum asÖ ½' L ( 3 7 Ó ' Ï 'Ð Õ 4 Ö ½' L ( 3 7 Ó ':7 Ï 'Ð Õ . Note that Í²/?Î ¦Í²/Z4 ) Î if and only if / a @ 4 a for some a .Thus /4=ÍT/?Î and / ; Í²/?Î each increase by 1 except at/ a @ 4 a , where the former skips from a @ 4 3 a toa @ 4 3 a 4 3 and the latter repeats the value a @ . Thusthe sums are½Ä' L ( 3 7P' ; ½

Ä\ L ( 3 7P\ Ò 4 ½

Ä' L ³ 3 7P' 4 ½

Ä\ L ( 3 7P\ Ò 3 4 )=- >

B–4 For a rational number × +.Ø expressed in lowest terms,define its height Ù × +.ØM to be

× 4 Ø: . Then forany × +.Øj expressed in lowest terms, we haveÙ G × +.ØMR^ Ø @ ; × @ 4 × Ø: ; since by assumption× and

Øare nonzero integers with

× ¦* Ø: , we haveÙ G × +CØMR; Ù × +CØM¾ Ø @ ; × @ 4 × Ø:; × ; Ø:º- 4 × Ø:;= × [;= Ø: × [;), Ø:;)[ 4 3 º 3 >It follows that G Ó ' Õ ¾ consists solely of numbers ofheight strictly larger than 3 /4 3 , and henceÚ ½' L ( G Ó ' Õ ¾ Û >Note: many choices for the height function are possible:one can take Ù × +.ØM*^ÜMÝ| × ÔM Ø: , or Ù × +.ØM equalto the total number of prime factors of × and

Ø, and so

on. The key properties of the height function are thaton one hand, there are only finitely many rationals withheight below any finite bound, and on the other hand,the height function is a sufficiently “algebraic” functionof its argument that one can relate the heights of × +CØand G × +.ØM .

B–5 Note that Â H< Â « implies that Â Â HR< Â Â « Rand hence

H « from the given equation. That is, Â is

injective. Since Â is also continuous, Â is either strictlyincreasing or strictly decreasing. Moreover, Â cannottend to a finite limit ® as

H 4 , or else we’d haveÂ Â HP; Â H<*H , with the left side bounded andthe right side unbounded. Similarly, Â cannot tend to afinite limit as

HÞ;. Together with monotonicity,

this yields that Â is also surjective.

PickH ³ arbitrary, and define

H ' for all / eß recur-sively by

H '#( Â H ' for /0 _ , andH ':79( Â 7#( H ' for /jà _ . Let n¹( Þ 4 d @ 4 ¢R+ 3

and n @ "; d @ 4 ¢R+ 3 and n @ be the roots ofH @ ;HX;e¥£_ , so that n ( 0 _ 0án,@ and) 0 nM( 0 n @ . Then there exist âÉ( â @ gã such thatH ' â ( n ' ( 4â@,n '@ for all / ^ß .

Suppose Â is strictly increasing. If â,@ ¦ä_ for somechoice of

H ³ , thenH ' is dominated by n '@ for / suffi-

ciently negative. But takingH ' and

H ' @ for / suffi-ciently negative of the right parity, we get

_ à H '=àH ' @ but Â H ' 0%Â H ' @ , contradiction. Thus â @ _; since

H ³ âÉ( andH ( âÉ(nM( , we have Â H n¹( H

for allH

. Analogously, if Â is strictly decreasing, thenâ@ e_ or elseH ' is dominated by n ' ( for / sufficiently

positive. But takingH ' and

H ' @ for / sufficiently pos-itive of the right parity, we get

_ à H ' @Ëà H ' butÂ H ' @ àÂ H ' , contradiction. Thus in that case,Â HP¾ n,@ H for allH

.

B–6 Yes, there must exist infinitely many such / . Let

bethe convex hull of the set of points

/ ' for / º_ .Geometrically,

is the intersection of all convex sets

(or even all halfplanes) containing the points / ' ;

algebraically,

is the set of pointsH# « which can

be written as â ( / ( ':å 4 Y,Y[Y 4â K / K 'æ for someâ[( >[>,> â K which are nonnegative of sum 1.

We prove that for infinitely many / , / ' is a vertexon the upper boundary of

, and that these / satisfy the

given condition. The condition that / ' is a vertex

on the upper boundary of

is equivalent to the exis-tence of a line passing through

/ ' with all otherpoints of

below it. That is, there should exist a 0 _

such that K à ' 4 a N ; / ç N ºe) > (1)

We first show that / ä) satisfies (1). The condition K + N `_ as N ä implies that K ; ( +: N ;m),_

as well. Thus the setf K ;5 ( R+8 N ;), i has an upper

bound a , and now K Ã= ( 4 a N ;), , as desired.

Next, we show that given one / satisfying (1), thereexists a larger one also satisfying (1). Again, the con-dition

K + N _as N

implies that K ; ' R+: N ; / Z _ as N . Thus the sequencef K ;Ì ' +: N ; / i K.è ' has a maximum element; sup-

pose N n is the largest value of N that achieves thismaximum, and put a hCé!;¥ ' +: n ; / . Then theline through

n é, of slope a lies strictly above N K

for N 0*n and passes through or lies above N K for


N àhn . Thus (1) holds for / n with a replaced bya ;Ëê for suitably smallê 0 _ .

By induction, we have that (1) holds for infinitely many/ . For any such / there exists a 0 _ such that

for Q ë). >,>[> / ;h) , the points / ; Q '87 O and /4mQ ' O lie below the line through / ' of slopea . That means

' O à ' 4 a Q and '87 O à ' ; a Q ;

adding these together gives ':7 O 4 ' O à 3 ' , as de-

sired.


Solutions to the 63rd William Lowell Putnam Mathematical CompetitionSaturday, December 7, 2002

Kiran Kedlaya and Lenny Ng

A1 By differentiating Pn(x)/(xk − 1)n+1, we find thatPn+1(x) = (xk−1)P ′n(x)− (n+1)kxk−1Pn(x); sub-stituting x = 1 yields Pn+1(1) = −(n + 1)kPn(1).SinceP0(1) = 1, an easy induction givesPn(1) =(−k)nn! for all n ≥ 0.

Note: one can also argue by expanding in Taylor seriesaround1. Namely, we have

1xk − 1

=1

k(x− 1) + · · ·=

1k

(x− 1)−1 + · · · ,

so

dn

dxn

1xk − 1

=(−1)nn!

k(x− 1)−n−1

and

Pn(x) = (xk − 1)n+1 dn

dxn

1xk − 1

= (k(x− 1) + · · · )n+1((−1)nn!

k(x− 1)−n−1 + · · ·

)= (−k)nn! + · · · .

A2 Draw a great circle through two of the points. There aretwo closed hemispheres with this great circle as bound-ary, and each of the other three points lies in one ofthem. By the pigeonhole principle, two of those threepoints lie in the same hemisphere, and that hemispherethus contains four of the five given points.

Note: by a similar argument, one can prove that amonganyn+3 points on ann-dimensional sphere, somen+2of them lie on a closed hemisphere. (One cannot get bywith only n+2 points: put them at the vertices of a reg-ular simplex.) Namely, anyn of the points lie on a greatsphere, which forms the boundary of two hemispheres;of the remaining three points, some two lie in the samehemisphere.

A3 Note that each of the sets1, 2, . . . , n has thedesired property. Moreover, for each setS with in-teger averagem that does not containm, S ∪ malso has averagem, while for each setT of more thanone element with integer averagem that containsm,T \m also has averagem. Thus the subsets other than1, 2, . . . , n can be grouped in pairs, soTn −n iseven.

A4 (partly due to David Savitt) Player 0 wins with opti-mal play. In fact, we prove that Player 1 cannot preventPlayer 0 from creating a row of all zeroes, a column ofall zeroes, or a2 × 2 submatrix of all zeroes. Each ofthese forces the determinant of the matrix to be zero.

For i, j = 1, 2, 3, let Aij denote the position in rowiand columnj. Without loss of generality, we may as-sume that Player 1’s first move is atA11. Player 0 thenplays atA22: 1 ∗ ∗

∗ 0 ∗∗ ∗ ∗

After Player 1’s second move, at least one ofA23 andA32 remains vacant. Without loss of generality, assumeA23 remains vacant; Player 0 then plays there.

After Player 1’s third move, Player 0 wins by playing atA21 if that position is unoccupied. So assume insteadthat Player 1 has played there. Thus of Player 1’s threemoves so far, two are atA11 andA21. Hence fori equalto one of 1 or 3, and forj equal to one of 2 or 3, thefollowing are both true:

(a) The2 × 2 submatrix formed by rows 2 andi andby columns 2 and 3 contains two zeroes and twoempty positions.

(b) Columnj contains one zero and two empty posi-tions.

Player 0 next plays atAij . To prevent a zero column,Player 1 must play in columnj, upon which Player 0completes the2× 2 submatrix in (a) for the win.

Note: one can also solve this problem directly by mak-ing a tree of possible play sequences. This tree can beconsiderably collapsed using symmetries: the symme-try between rows and columns, the invariance of theoutcome under reordering of rows or columns, and thefact that the scenario after a sequence of moves doesnot depend on the order of the moves (sometimes called“transposition invariance”).

Note (due to Paul Cheng): one can reduce Determi-nant Tic-Tac-Toe to a variant of ordinary tic-tac-toe.Namely, consider a tic-tac-toe grid labeled as follows:

A11 A22 A33

A23 A31 A12

A32 A13 A21


Then each term in the expansion of the determinant oc-curs in a row or column of the grid. Suppose Player1 first plays in the top left. Player 0 wins by playingfirst in the top row, and second in the left column. Thenthere are only one row and column left for Player 1 tothreaten, and Player 1 cannot already threaten both onthe third move, so Player 0 has time to block both.

A5 It suffices to prove that for any relatively prime positiveintegersr, s, there exists an integern with an = r andan+1 = s. We prove this by induction onr+s, the caser + s = 2 following from the fact thata0 = a1 = 1.Givenr ands not both 1 withgcd(r, s) = 1, we musthaver 6= s. If r > s, then by the induction hypothesiswe havean = r − s andan+1 = s for somen; thena2n+2 = r anda2n+3 = s. If r < s, then we havean = r andan+1 = s− r for somen; thena2n+1 = randa2n+2 = s.

Note: a related problem is as follows. Starting with thesequence

01,10,

repeat the following operation: insert between each pairab and c

d the paira+cb+d . Prove that each positive rational

number eventually appears.

Observe that by induction, ifab and cd are consecutive

terms in the sequence, thenbc − ad = 1. The sameholds for consecutive terms of then-th Farey sequence,the sequence of rational numbers in[0, 1] with denomi-nator (in lowest terms) at mostn.

A6 The sum converges forb = 2 and diverges forb ≥ 3.We first considerb ≥ 3. Suppose the sum converges;then the fact thatf(n) = nf(d) wheneverbd−1 ≤ n ≤bd − 1 yields

∞∑n=1

1f(n)

=∞∑

d=1

1f(d)

bd−1∑n=bd−1

1n

. (1)

However, by comparing the integral of1/x with a Rie-mann sum, we see that

bd−1∑n=bd−1

1n

>

∫ bd

bd−1

dx

x

= log(bd)− log(bd−1) = log b,

wherelog denotes the natural logarithm. Thus (1) yields

∞∑n=1

1f(n)

> (log b)∞∑

n=1

1f(n)

,

a contradiction sincelog b > 1 for b ≥ 3. Therefore thesum diverges.

For b = 2, we have a slightly different identity becausef(2) 6= 2f(2). Instead, for any positive integeri, wehave

2i−1∑n=1

1f(n)

= 1 +12

+16

+i∑

d=3

1f(d)

2d−1∑n=2d−1

1n

. (2)

Again comparing an integral to a Riemann sum, we seethat ford ≥ 3,

2d−1∑n=2d−1

1n

<1

2d−1− 1

2d+

∫ 2d

2d−1

dx

x

=12d

+ log 2

≤ 18

+ log 2 < 0.125 + 0.7 < 1.

Putc = 18 +log 2 andL = 1+ 1

2 + 16(1−c) . Then we can

prove that∑2i−1

n=11

f(n) < L for all i ≥ 2 by inductionon i. The casei = 2 is clear. For the induction, notethat by (2),

2i−1∑n=1

1f(n)

< 1 +12

+16

+ ci∑

d=3

1f(d)

< 1 +12

+16

+ c1

6(1− c)

= 1 +12

+1

6(1− c)= L,

as desired. We conclude that∑∞

n=11

f(n) converges to alimit less than or equal toL.

Note: the above argument proves that the sum forb = 2is at mostL < 2.417. One can also obtain a lowerbound by the same technique, namely1 + 1

2 + 16(1−c′)

with c′ = log 2. This bound exceeds2.043. (By con-trast, summing the first 100000 terms of the series onlyyields a lower bound of1.906.) Repeating the same ar-guments withd ≥ 4 as the cutoff yields the upper bound2.185 and the lower bound2.079.

B1 The probability is1/99. In fact, we show by induc-tion on n that aftern shots, the probability of havingmade any number of shots from1 to n − 1 is equal to1/(n − 1). This is evident forn = 2. Given the resultfor n, we see that the probability of makingi shots aftern + 1 attempts is

i− 1n

1n− 1

+(

1− i

n

)1

n− 1=

(i− 1) + (n− i)n(n− 1)

=1n

,

as claimed.

2


B2 (Note: the problem statement assumes that all polyhe-dra are connected and that no two edges share more thanone face, so we will do likewise. In particular, these aretrue for all convex polyhedra.) We show that in factthe first player can win on the third move. Suppose thepolyhedron has a faceA with at least four edges. If thefirst player plays there first, after the second player’sfirst move there will be three consecutive facesB,C,Dadjacent toA which are all unoccupied. The first playerwins by playing inC; after the second player’s secondmove, at least one ofB andD remains unoccupied, andeither is a winning move for the first player.

It remains to show that the polyhedron has a face with atleast four edges. (Thanks to Russ Mann for suggestingthe following argument.) Suppose on the contrary thateach face has only three edges. Starting with any faceF1 with verticesv1, v2, v3, let v4 be the other endpointof the third edge out ofv1. Then the faces adjacent toF1

must have verticesv1, v2, v4; v1, v3, v4; andv2, v3, v4.Thus v1, v2, v3, v4 form a polyhedron by themselves,contradicting the fact that the given polyhedron is con-nected and has at least five vertices. (One can also de-duce this using Euler’s formulaV − E + F = 2 − 2g,whereV,E, F are the numbers of vertices, edges andfaces, respectively, andg is the genus of the polyhe-dron. For a convex polyhedron,g = 0 and you get the“usual” Euler’s formula.)

Note: Walter Stromquist points out the following coun-terexample if one relaxes the assumption that a pair offaces may not share multiple edges. Take a tetrahedronand remove a smaller tetrahedron from the center of anedge; this creates two small triangular faces and turnstwo of the original faces into hexagons. Then the sec-ond player can draw by signing one of the hexagons,one of the large triangles, and one of the small trian-gles. (He does this by “mirroring”: wherever the firstplayer signs, the second player signs the other face ofthe same type.)

B3 The desired inequalities can be rewritten as

1− 1n

< exp(

1 + n log(

1− 1n

))< 1− 1

2n.

By taking logarithms, we can rewrite the desired in-equalities as

− log(

1− 12n

)< −1− n log

(1− 1

n

)< − log

(1− 1

n

).

Rewriting these in terms of the Taylor expansion of− log(1−x), we see that the desired result is also equiv-alent to

∞∑i=1

1i2ini

<∞∑

i=1

1(i + 1)ni

<∞∑

i=1

1ini

,

which is evident because the inequalities hold term byterm.

Note: David Savitt points out that the upper bound canbe improved from1/(ne) to 2/(3ne) with a slightlymore complicated argument. (In fact, for anyc > 1/2,one has an upper bound ofc/(ne), but only forn abovea certain bound depending onc.)

B4 Use the following strategy: guess1, 3, 4, 6, 7, 9, . . . un-til the target numbern is revealed to be equal to or lowerthan one of these guesses. Ifn ≡ 1 (mod 3), it will beguessed on an odd turn. Ifn ≡ 0 (mod 3), it will beguessed on an even turn. Ifn ≡ 2 (mod 3), thenn + 1will be guessed on an even turn, forcing a guess ofn onthe next turn. Thus the probability of success with thisstrategy is1335/2002 > 2/3.

Note: for any positive integerm, this strategy winswhen the number is being guessed from[1,m] withprobability 1

mb 2m+13 c. We can prove that this is best

possible as follows. Letam denotem times the proba-bility of winning when playing optimally. Also, letbm

denotem times the corresponding probability of win-ning if the objective is to select the number in an evennumber of guesses instead. (For definiteness, extend thedefinitions to incorporatea0 = 0 andb0 = 0.)

We first claim thatam = 1+max1≤k≤mbk−1+bm−kandbm = max1≤k≤mak−1 + am−k for m ≥ 1. Toestablish the first recursive identity, suppose that ourfirst guess is some integerk. We automatically win ifn = k, with probability1/m. If n < k, with proba-bility (k − 1)/m, then we wish to guess an integer in[1, k − 1] in an even number of guesses; the probabil-ity of success when playing optimally isbk−1/(k − 1),by assumption. Similarly, ifn < k, with probability(m−k)/m, then the subsequent probability of winningis bm−k/(m − k). In sum, the overall probability ofwinning if k is our first guess is(1 + bk−1 + bm−k)/m.For optimal strategy, we choosek such that this quan-tity is maximized. (Note that this argument still holdsif k = 1 or k = m, by our definitions ofa0 andb0.)The first recursion follows, and the second recursion isestablished similarly.

We now prove by induction thatam = b(2m + 1)/3candbm = b2m/3c for m ≥ 0. The inductive step relieson the inequalitybxc + byc ≤ bx + yc, with equal-ity when one ofx, y is an integer. Now suppose thatai = b(2i + 1)/3c andbi = b2i/3c for i < m. Then

1 + bk−1 + bm−k = 1 +⌊

2(k − 1)3

⌋+

⌊2(m− k)

3

⌋≤

⌊2m

3

⌋and similarlyak−1 + am−k ≤ b(2m + 1)/3c, withequality in both cases attained, e.g., whenk = 1. Theinductive formula foram andbm follows.

3


B5 (due to Dan Bernstein) PutN = 2002!. Then ford = 1, . . . , 2002, the numberN2 written in baseb =N/d − 1 has digitsd2, 2d2, d2. (Note that these reallyare digits because2(2002)2 < (2002!)2/2002− 1.)

Note: one can also produce an integerN which has baseb digits 1, ∗, 1 for n different values ofb, as follows.Choosec with 0 < c < 21/n. For m a large positiveinteger, putN = 1+(m+1) · · · (m+n)bcmcn−2. Form sufficiently large, the bases

b =N − 1

(m + i)mn−2=

∏j 6=i

(m + j)

for i = 1, . . . , n will have the properties thatN ≡ 1(mod b) andb2 < N < 2b2 for m sufficiently large.

Note (due to Russ Mann): one can also give a “noncon-structive” argument. LetN be a large positive integer.For b ∈ (N2, N3), the number of 3-digit base-b palin-dromes in the range[b2, N6 − 1] is at least⌊

N6 − b2

b

⌋− 1 ≥ N6

b2− b− 2,

since there is a palindrome in each interval[kb, (k +1)b − 1] for k = b, . . . , b2 − 1. Thus the average num-ber of bases for which a number in[1, N6 − 1] is atleast

1N6

N3−1∑b=N2+1

(N6

b− b− 2

)≥ log(N)− c

for some constantc > 0. TakeN so that the right sideexceeds2002; then at least one number in[1, N6 − 1]is a base-b palindrome for at least 2002 values ofb.

B6 We prove that the determinant is congruent modulop to

x

p−1∏i=0

(y + ix)p−1∏

i,j=0

(z + ix + jy). (3)

We first check that

p−1∏i=0

(y + ix) ≡ yp − xp−1y (mod p). (4)

Since both sides are homogeneous as polynomials inxandy, it suffices to check (4) forx = 1, as a congru-ence between polynomials. Now note that the right sidehas0, 1, . . . , p − 1 as roots modulop, as does the left

side. Moreover, both sides have the same leading coef-ficient. Since they both have degree onlyp, they mustthen coincide.

We thus have

x

p−1∏i=0

(y + ix)p−1∏

i,j=0

(z + ix + jy)

≡ x(yp − xp−1y)p−1∏j=0

((z + jy)p − xp−1(z + jy))

≡ (xyp − xpy)p−1∏j=0

(zp − xp−1z + jyp − jxp−1y)

≡ (xyp − xpy)((zp − xp−1z)p

− (yp − xp−1y)p−1(zp − xp−1z))

≡ (xyp − xpy)(zp2− xp2−pzp)

− x(yp − xp−1y)p(zp − xp−1z)

≡ xypzp2− xpyzp2

− xp2−p+1ypzp + xp2yzp

− xyp2zp + xp2−p+1ypzp + xpyp2

z − xp2ypz

≡ xypzp2+ yzpxp2

+ zxpyp2

− xzpyp2− yxpzp2

− zypxp2,

which is precisely the desired determinant.

Note: a simpler conceptual proof is as follows. (Ev-erything in this paragraph will be modulop.) Notethat for any integersa, b, c, the column vector[ax +by + cz, (ax + by + cz)p, (ax + by + cz)p2

] is a linearcombination of the columns of the given matrix. Thusax+by+cz divides the determinant. In particular, all ofthe factors of (3) divide the determinant; since both (3)and the determinant have degreep2 + p + 1, they agreeup to a scalar multiple. Moreover, they have the samecoefficient ofzp2

ypx (since this term only appears inthe expansion of (3) when you choose the first term ineach factor). Thus the determinant is congruent to (3),as desired.

Either argument can be used to generalize to a corre-spondingn × n determinant, called a Moore determi-nant; we leave the precise formulation to the reader.Note the similarity with the classical Vandermonde de-terminant: ifA is then × n matrix with Aij = xj

i fori, j = 0, . . . , n− 1, then

det(A) =∏

1≤i<j≤n

(xj − xi).

4


Solutions to the 64th William Lowell Putnam Mathematical CompetitionSaturday, December 6, 2003


A1 There aren such sums. More precisely, there is exactlyone such sum withk terms for each ofk = 1, . . . , n(and clearly no others). To see this, note that ifn =a1 + a2 + · · ·+ ak with a1 ≤ a2 ≤ · · · ≤ ak ≤ a1 + 1,then

ka1 = a1 + a1 + · · ·+ a1

≤ n ≤ a1 + (a1 + 1) + · · ·+ (a1 + 1)= ka1 + k − 1.

However, there is a unique integera1 satisfying theseinequalities, namelya1 = bn/kc. Moreover, oncea1

is fixed, there arek different possibilities for the suma1 + a2 + · · · + ak: if i is the last integer such thatai = a1, then the sum equalska1 + (i − 1). The pos-sible values ofi are1, . . . , k, and exactly one of thesesums comes out equal ton, proving our claim.

Note: In summary, there is a unique partition ofn withk terms that is “as equally spaced as possible”. Onecan also obtain essentially the same construction induc-tively: except for the all-ones sum, each partition ofnis obtained by “augmenting” a unique partition ofn−1.

A2 First solution: Assume without loss of generality thatai + bi > 0 for eachi (otherwise both sides of the de-sired inequality are zero). Then the AM-GM inequalitygives (

a1 · · · an

(a1 + b1) · · · (an + bn)

)1/n

≤ 1n

(a1

a1 + b1+ · · ·+ an

an + bn

),

and likewise with the roles ofa andb reversed. Addingthese two inequalities and clearing denominators yieldsthe desired result.

Second solution: Write the desired inequality in theform

(a1+b1) · · · (an+bn) ≥ [(a1 · · · an)1/n+(b1 · · · bn)1/n]n,

expand both sides, and compare the terms on bothsides in which k of the terms are among theai. On the left, one has the product of eachk-element subset of1, . . . , n; on the right, one has(nk

)(a1 · · · an)k/n · · · (b1 . . . bn)(n−k)/n, which is pre-

cisely(nk

)times the geometric mean of the terms on

the left. Thus AM-GM shows that the terms under con-sideration on the left exceed those on the right; addingthese inequalities over allk yields the desired result.

Third solution: Since both sides are continuous in eachai, it is sufficient to prove the claim witha1, . . . , an allpositive (the general case follows by taking limits assome of theai tend to zero). Putri = bi/ai; then thegiven inequality is equivalent to

(1 + r1)1/n · · · (1 + rn)1/n ≥ 1 + (r1 · · · rn)1/n.

In terms of the function

f(x) = log(1 + ex)

and the quantitiessi = log ri, we can rewrite the de-sired inequality as

1n

(f(s1) + · · ·+ f(sn)) ≥ f

(s1 + · · ·+ sn

n

).

This will follow from Jensen’s inequality if we can ver-ify that f is a convex function; it is enough to check thatf ′′(x) > 0 for all x. In fact,

f ′(x) =ex

1 + ex= 1− 1

1 + ex

is an increasing function ofx, so f ′′(x) > 0 andJensen’s inequality thus yields the desired result. (Aslong as theai are all positive, equality holds whens1 = · · · = sn, i.e., when the vectors(a1, . . . , an) and(b1, . . . , bn). Of course other equality cases crop up ifsome of theai vanish, i.e., ifa1 = b1 = 0.)

Fourth solution: We apply induction onn, the casen = 1 being evident. First we verify the auxiliary in-equality

(an + bn)(cn + dn)n−1 ≥ (acn−1 + bdn−1)n

for a, b, c, d ≥ 0. The left side can be written as

ancn(n−1) + bndn(n−1)

+n−1∑i=1

(n− 1

i

)ancnidn(n−1−i)

+n−1∑i=1

(n− 1i− 1

)bncn(n−i)dn(i−1).


Applying the weighted AM-GM inequality betweenmatching terms in the two sums yields

(an + bn)(cn + dn)n−1

≥ ancn(n−1) + bndn(n−1)

+n−1∑i=1

(n

i

)aibn−ic(n−1)id(n−1)(n−i),

proving the auxiliary inequality.

Now given the auxiliary inequality and then−1 case ofthe desired inequality, we apply the auxiliary inequalitywith a = a

1/n1 , b = b

1/n1 , c = (a2 · · · an)1/n(n−1),

d = (b2 . . . bn)1/n(n−1). The right side will be then-thpower of the desired inequality. The left side comes outto

(a1 + b1)((a2 · · · an)1/(n−1) + (b2 · · · bn)1/(n−1))n−1,

and by the induction hypothesis, the second factor isless than(a2 + b2) · · · (an + bn). This yields the de-sired result.

Note: Equality holds if and only ifai = bi = 0 forsomei or if the vectors(a1, . . . , an) and (b1, . . . , bn)are proportional. As pointed out by Naoki Sato, theproblem also appeared on the 1992 Irish MathematicalOlympiad. It is also a special case of a classical in-equality, known as Holder’s inequality, which general-izes the Cauchy-Schwarz inequality (this is visible fromthen = 2 case); the first solution above is adapted fromthe standard proof of Holder’s inequality. We don’tknow whether the declaration “Apply Holder’s inequal-ity” by itself is considered an acceptable solution to thisproblem.

A3 First solution: Write

f(x) = sin x + cos x + tanx + cot x + sec x + csc x

= sinx + cos x +1

sinx cos x+

sinx + cos x

sinx cos x.

We can writesinx + cos x =√

2 cos(π/4 − x); thissuggests making the substitutiony = π/4 − x. In thisnew coordinate,

sinx cos x =12

sin 2x =12

cos 2y,

and writingc =√

2 cos y, we have

f(y) = (1 + c)(

1 +2

c2 − 1

)− 1

= c +2

c− 1.

We must analyze this function ofc in the range[−√

2,√

2]. Its value atc = −√

2 is 2−3√

2 < −2.24,and atc =

√2 is 2 + 3

√2 > 6.24. Its derivative is

1− 2/(c− 1)2, which vanishes when(c− 1)2 = 2, i.e.,wherec = 1 ±

√2. Only the valuec = 1 −

√2 is in

bounds, at which the value off is 1 − 2√

2 > −1.83.As for the pole atc = 1, we observe thatf decreasesas c approaches from below (so takes negative valuesfor all c < 1) and increases asc approaches from above(so takes positive values for allc > 1); from the datacollected so far, we see thatf has no sign crossings, sothe minimum of|f | is achieved at a critical point off .We conclude that the minimum of|f | is 2

√2− 1.

Alternate derivation (due to Zuming Feng): We can alsominimize|c + 2/(c− 1)| without calculus (or worryingabout boundary conditions). Forc > 1, we have

1 + (c− 1) +2

c− 1≥ 1 + 2

√2

by AM-GM on the last two terms, with equality forc − 1 =

√2 (which is out of range). Forc < 1, we

similarly have

−1 + 1− c +2

1− c≥ −1 + 2

√2,

here with equality for1− c =√

2.

Second solution:Write

f(a, b) = a + b +1ab

+a + b

ab.

Then the problem is to minimize|f(a, b)| subject to theconstrainta2 + b2 − 1 = 0. Since the constraint re-gion has no boundary, it is enough to check the valueat each critical point and each potential discontinuity(i.e., whereab = 0) and select the smallest value (afterchecking thatf has no sign crossings).

We locate the critical points using the Lagrange mul-tiplier condition: the gradient off should be parallelto that of the constraint, which is to say, to the vector(a, b). Since

∂f

∂a= 1− 1

a2b− 1

a2

and similarly forb, the proportionality yields

a2b3 − a3b2 + a3 − b3 + a2 − b2 = 0.

The irreducible factors of the left side are1 + a, 1 + b,a− b, andab− a− b. So we must check what happenswhen any of those factors, ora or b, vanishes.

If 1 + a = 0, thenb = 0, and the singularity off be-comes removable when restricted to the circle. Namely,we have

f = a + b +1a

+b + 1ab

anda2+b2−1 = 0 implies(1+b)/a = a/(1−b). Thuswe havef = −2; the same occurs when1 + b = 0.

2


If a − b = 0, then a = b = ±√

2/2 and eitherf = 2 + 3

√2 > 6.24, or f = 2− 3

√2 < −2.24.

If a = 0, then eitherb = −1 as discussed above, orb = 1. In the latter case,f blows up as one approachesthis point, so there cannot be a global minimum there.

Finally, if ab− a− b = 0, then

a2b2 = (a + b)2 = 2ab + 1

and soab = 1±√

2. The plus sign is impossible since|ab| ≤ 1, soab = 1−

√2 and

f(a, b) = ab +1ab

+ 1

= 1− 2√

2 > −1.83.

This yields the smallest value of|f | in the list (and in-deed no sign crossings are possible), so2

√2− 1 is the

desired minimum of|f |.Note: Instead of using the geometry of the graph offto rule out sign crossings, one can verify explicitly thatf cannot take the value 0. In the first solution, note thatc + 2/(c − 1) = 0 impliesc2 − c + 2 = 0, which hasno real roots. In the second solution, we would have

a2b + ab2 + a + b = −1.

Squaring both sides and simplifying yields

2a3b3 + 5a2b2 + 4ab = 0,

whose only real root isab = 0. But the cases withab = 0 do not yieldf = 0, as verified above.

A4 We split into three cases. Note first that|A| ≥ |a|, byapplying the condition for largex.

Case 1:B2 − 4AC > 0. In this caseAx2 + Bx + Chas two distinct real rootsr1 andr2. The condition im-plies thatax2 + bx + c also vanishes atr1 andr2, sob2 − 4ac > 0. Now

B2 − 4AC = A2(r1 − r2)2

≥ a2(r1 − r2)2

= b2 − 4ac.

Case 2: B2 − 4AC ≤ 0 and b2 − 4ac ≤ 0. As-sume without loss of generality thatA ≥ a > 0, andthat B = 0 (by shifting x). ThenAx2 + Bx + C ≥ax2 + bx + c ≥ 0 for all x; in particular,C ≥ c ≥ 0.Thus

4AC −B2 = 4AC

≥ 4ac

≥ 4ac− b2.

Alternate derivation (due to Robin Chapman): the el-lipse Ax2 + Bxy + Cy2 = 1 is contained within the

ellipseax2 + bxy + cy2 = 1, and their respective en-closed areas areπ/(4AC −B2) andπ/(4ac− b2).

Case 3: B2 − 4AC ≤ 0 and b2 − 4ac > 0. SinceAx2 + Bx + C has a graph not crossing thex-axis, sodo (Ax2 + Bx + C)± (ax2 + bx + c). Thus

(B − b)2 − 4(A− a)(C − c) ≤ 0,

(B + b)2 − 4(A + a)(C + c) ≤ 0

and adding these together yields

2(B2 − 4AC) + 2(b2 − 4ac) ≤ 0.

Henceb2 − 4ac ≤ 4AC −B2, as desired.

A5 First solution: We represent a Dyckn-path by a se-quencea1 · · · a2n, where eachai is either (1, 1) or(1,−1).

Given an (n − 1)-path P = a1 · · · a2n−2, wedistinguish two cases. IfP has no returnsof even-length, then letf(P ) denote then-path(1, 1)(1,−1)P . Otherwise, letaiai+1 · · · aj denote therightmost even-length return inP , and let f(P ) =(1, 1)a1a2 · · · aj(1,−1)aj+1 · · · a2n−2. Thenf clearlymaps the set of Dyck(n − 1)-paths to the set of Dyckn-paths having no even return.

We claim thatf is bijective; to see this, we simplyconstruct the inverse mapping. Given ann-pathP , letR = aiai+1...aj denote the leftmost return inP , andlet g(P ) denote the path obtained by removinga1 andaj from P . Then evidentlyf g andg f are identitymaps, proving the claim.

Second solution: (by Dan Bernstein) LetCn be thenumber of Dyck paths of lengthn, let On be the num-ber of Dyck paths whose final return has odd length,and letXn be the number of Dyck paths with no returnof even length.

We first exhibit a recursion forOn; note thatO0 = 0.Given a Dyckn-path whose final return has odd length,split it just after its next-to-last return. For somek (pos-sibly zero), this yields a Dyckk-path, an upstep, a Dyck(n−k− 1)-path whose odd return has even length, anda downstep. Thus forn ≥ 1,

On =n−1∑k=0

Ck(Cn−k−1 −On−k−1).

We next exhibit a similar recursion forXn; note thatX0 = 1. Given a Dyckn-path with no even return,splitting as above yields for somek a Dyckk-path withno even return, an upstep, a Dyck(n−k−1)-path whosefinal return has even length, then a downstep. Thus forn ≥ 1,

Xn =n−1∑k=0

Xk(Cn−k−1 −On−k−1).

3


To conclude, we verify thatXn = Cn−1 for n ≥ 1,by induction onn. This is clear forn = 1 sinceX1 = C0 = 1. Given Xk = Ck−1 for k < n, wehave

Xn =n−1∑k=0

Xk(Cn−k−1 −On−k−1)

= Cn−1 −On−1 +n−1∑k=1

Ck−1(Cn−k−1 −On−k−1)

= Cn−1 −On−1 + On−1

= Cn−1,

as desired.

Note: Since the problem only asked about theexistenceof a one-to-one correspondence, we believe that anyproof, bijective or not, that the two sets have the samecardinality is an acceptable solution. (Indeed, it wouldbe highly unusual to insist on using or not using a spe-cific proof technique!) The second solution above canalso be phrased in terms of generating functions. Also,the Cn are well-known to equal the Catalan numbers

1n+1

(2nn

); the problem at hand is part of a famous exer-

cise in Richard Stanley’sEnumerative Combinatorics,Volume 1giving 66 combinatorial interpretations of theCatalan numbers.

A6 First solution: Yes, such a partition is possible. Toachieve it, place each integer intoA if it has an evennumber of 1s in its binary representation, and intoBif it has an odd number. (One discovers this by simplyattempting to place the first few numbers by hand andnoticing the resulting pattern.)

To show thatrA(n) = rB(n), we exhibit a bijection be-tween the pairs(a1, a2) of distinct elements ofA witha1 + a2 = n and the pairs(b1, b2) of distinct elementsof B with b1 + b2 = n. Namely, given a pair(a1, a2)with a1 + a2 = n, write both numbers in binary andfind the lowest-order place in which they differ (such aplace exists becausea1 6= a2). Change both numbersin that place and call the resulting numbersb1, b2. Thena1 + a2 = b1 + b2 = n, but the parity of the number of1s inb1 is opposite that ofa1, and likewise betweenb2

anda2. This yields the desired bijection.

Second solution: (by Micah Smukler) Writeb(n) forthe number of 1s in the base 2 expansion ofn, andf(n) = (−1)b(n). Then the desired partition can bedescribed asA = f−1(1) andB = f−1(−1). Sincef(2n) + f(2n + 1) = 0, we have

n∑i=0

f(n) =

0 n oddf(n) n even.

If p, q are both inA, thenf(p) + f(q) = 2; if p, q areboth inB, thenf(p)+f(q) = −2; if p, q are in different

sets, thenf(p) + f(q) = 0. In other words,

2(rA(n)− rB(n)) =∑

p+q=n,p<q

(f(p) + f(q))

and it suffices to show that the sum on the right is alwayszero. Ifn is odd, that sum is visibly

∑ni=0 f(i) = 0. If

n is even, the sum equals(n∑

i=0

f(i)

)− f(n/2) = f(n)− f(n/2) = 0.

This yields the desired result.

Third solution: (by Dan Bernstein) Putf(x) =∑n∈A xn and g(x) =

∑n∈B xn; then the value of

rA(n) (resp.rB(n)) is the coefficient ofxn in f(x)2 −f(x2) (resp.g(x)2−g(x2)). From the evident identities

11− x

= f(x) + g(x)

f(x) = f(x2) + xg(x2)

g(x) = g(x2) + xf(x2),

we have

f(x)− g(x) = f(x2)− g(x2) + xg(x2)− xf(x2)

= (1− x)(f(x2)− g(x2))

=f(x2)− g(x2)f(x) + g(x)

.

We deduce thatf(x)2− g(x)2 = f(x2)− g(x2), yield-ing the desired equality.

Note: This partition is actually unique, up to inter-changingA andB. More precisely, the condition that0 ∈ A andrA(n) = rB(n) for n = 1, . . . ,m uniquelydetermines the positions of0, . . . ,m. We see this byinduction onm: given the result form − 1, switchingthe location ofm changesrA(m) by one and does notchangerB(m), so it is not possible for both positions towork. Robin Chapman points out this problem is solvedin D.J. Newman’sAnalytic Number Theory(Springer,1998); in that solution, one uses generating functions tofind the partition and establish its uniqueness, not justverify it.

B1 No, there do not.

First solution: Suppose the contrary. By settingy =−1, 0, 1 in succession, we see that the polynomials1 − x + x2, 1, 1 + x + x2 are linear combinations ofa(x) andb(x). But these three polynomials are linearlyindependent, so cannot all be written as linear combina-tions of two other polynomials, contradiction.

Alternate formulation: the given equation expresses adiagonal matrix with1, 1, 1 and zeroes on the diagonal,which has rank 3, as the sum of two matrices of rank 1.But the rank of a sum of matrices is at most the sum ofthe ranks of the individual matrices.

4


Second solution: It is equivalent (by relabeling andrescaling) to show that1+xy +x2y2 cannot be writtenasa(x)d(y)− b(x)c(y). Write a(x) =

∑aix

i, b(x) =∑bix

i, c(y) =∑

cjyj , d(y) =

∑djy

j . We now startcomparing coefficients of1+xy+x2y2. By comparingcoefficients of1+xy +x2y2 anda(x)d(y)− b(x)c(y),we get

1 = aidi − bici (i = 0, 1, 2)0 = aidj − bicj (i 6= j).

The first equation says thatai andbi cannot both vanish,andci anddi cannot both vanish. The second equationsays thatai/bi = cj/dj wheni 6= j, where both sidesshould be viewed inR ∪ ∞ (and neither is undeter-mined if i, j ∈ 0, 1, 2). But then

a0/b0 = c1/d1 = a2/b2 = c0/d0

contradicting the equationa0d0 − b0c0 = 1.

Third solution: We work over the complex numbers,in which we have a primitive cube rootω of 1. We alsouse without further comment unique factorization forpolynomials in two variables over a field. And we keepthe relabeling of the second solution.

Suppose the contrary. Since1 + xy + x2y2 = (1 −xy/ω)(1−xy/ω2), the rational functiona(ω/y)d(y)−b(ω/y)c(y) must vanish identically (that is, coefficientby coefficient). If one of the polynomials, saya, van-ished identically, then one ofb or c would also, and thedesired inequality could not hold. So none of them van-ish identically, and we can write

c(y)d(y)

=a(ω/y)b(ω/y)

.

Likewise,

c(y)d(y)

=a(ω2/y)b(ω2/y)

.

Put f(x) = a(x)/b(x); then we havef(ωx) = f(x)identically. That is,a(x)b(ωx) = b(x)a(ωx). Sinceaandb have no common factor (otherwise1+xy +x2y2

would have a factor divisible only byx, which it doesn’tsince it doesn’t vanish identically for any particularx),a(x) dividesa(ωx). Since they have the same degree,they are equal up to scalars. It follows that one ofa(x), xa(x), x2a(x) is a polynomial inx3 alone, andlikewise forb (with the same power ofx).

If xa(x) andxb(x), or x2a(x) andx2b(x), are poly-nomials inx3, thena andb are divisible byx, but weknow a andb have no common factor. Hencea(x) andb(x) are polynomials inx3. Likewise, c(y) andd(y)are polynomials iny3. But then1 + xy + x2y2 =a(x)d(y)− b(x)c(y) is a polynomial inx3 andy3, con-tradiction.

Note: The third solution only works over fields of char-acteristic not equal to 3, whereas the other two workover arbitrary fields. (In the first solution, one must re-place−1 by another value if working in characteristic2.)

B2 It is easy to see by induction that thej-th entry ofthek-th sequence (where the original sequence isk =1) is

∑ki=1

(k−1i−1

)/(2k−1(i + j − 1)), and soxn =

12n−1

∑ni=1

(n−1i−1

)/i. Now

(n−1i−1

)/i =

(ni

)/n; hence

xn =1

n2n−1

n∑i=1

(n

i

)=

2n − 1n2n−1

< 2/n,

as desired.

B3 First solution: It is enough to show that for each primep, the exponent ofp in the prime factorization of bothsides is the same. On the left side, it is well-known thatthe exponent ofp in the prime factorization ofn! is

n∑i=1

⌊n

pi

⌋.

(To see this, note that thei-th term counts the multiplesof pi among1, . . . , n, so that a number divisible exactlyby pi gets counted exactlyi times.) This number can bereinterpreted as the cardinality of the setS of points inthe plane with positive integer coordinates lying on orunder the curvey = np−x: namely, each summand isthe number of points ofS with x = i.

On the right side, the exponent ofp in the primefactorization of lcm(1, . . . , bn/ic) is blogpbn/icc =blogp(n/i)c. However, this is precisely the number ofpoints ofS with y = i. Thus

n∑i=1

blogpbn/icc =n∑

i=1

⌊n

pi

⌋,

and the desired result follows.

Second solution:We prove the result by induction onn, the casen = 1 being obvious. What we actuallyshow is that going fromn − 1 to n changes both sidesby the same multiplicative factor, that is,

n =n−1∏i=1

lcm1, 2, . . . , bn/iclcm1, 2, . . . , b(n− 1)/ic

.

Note that thei-th term in the product is equal to 1 ifn/iis not an integer, i.e., ifn/i is not a divisor ofn. It isalso equal to 1 ifn/i is a divisor ofn but not a primepower, since any composite number divides the lcm ofall smaller numbers. However, ifn/i is a power ofp,then thei-th term is equal top.

Sincen/i runs over all proper divisors ofn, the producton the right side includes one factor of the primep foreach factor ofp in the prime factorization ofn. Thusthe whole product is indeed equal ton, completing theinduction.

5


B4 First solution: Putg = r1 +r2, h = r3 +r4, u = r1r2,v = r3r4. We are given thatg is rational. The followingare also rational:

−b

a= g + h

c

a= gh + u + v

−d

a= gv + hu

From the first line,h is rational. From the second line,u + v is rational. From the third line,g(u + v)− (gv +hu) = (g − h)u is rational. Sinceg 6= h, u is rational,as desired.

Second solution:This solution uses some basic Galoistheory. We may assumer1 6= r2, since otherwise theyare both rational and so then isr1r2.

Let τ be an automorphism of the field of algebraic num-bers; thenτ maps eachri to another one, and fixesthe rational numberr1 + r2. If τ(r1) equals one ofr1 or r2, thenτ(r2) must equal the other one, and viceversa. Thusτ either fixes the setr1, r2 or moves itto r3, r4. But if the latter happened, we would haver1 + r2 = r3 + r4, contrary to hypothesis. Thusτ fixesthe setr1, r2 and in particular the numberr1r2. Sincethis is true for anyτ , r1r2 must be rational.

Note: The conclusion fails if we allowr1+r2 = r3+r4.For instance, take the polynomialx4 − 2 and labelits roots so that(x − r1)(x − r2) = x2 −

√2 and

(x− r3)(x− r4) = x2 +√

2.

B5 Place the unit circle on the complex plane so thatA,B,C correspond to the complex numbers1, ω, ω2,whereω = e2πi/3, and letP correspond to the complexnumberx. The distancesa, b, c are then|x − 1|, |x −ω|, |x− ω2|. Now the identity

(x− 1) + ω(x− ω) + ω2(x− ω2) = 0

implies that there is a triangle whose sides, as vec-tors, correspond to the complex numbersx − 1, ω(x −ω), ω2(x− ω2); this triangle has sides of lengtha, b, c.

To calculate the area of this triangle, we first note a moregeneral formula. If a triangle in the plane has verticesat 0, v1 = s1 + it1, v2 = s2 + it2, then it is wellknown that the area of the triangle is|s1t2 − s2t1|/2 =|v1v2 − v2v1|/4. In our case, we havev1 = x − 1 andv2 = ω(x− ω); then

v1v2 − v2v1 = (ω2 − ω)(xx− 1) = i√

3(|x|2 − 1).

Hence the area of the triangle is√

3(1− |x|2)/4, whichdepends only on the distance|x| from P to O.

B6 First solution: (composite of solutions by Feng Xieand David Pritchard) Letµ denote Lebesgue measure

on [0, 1]. Define

E+ = x ∈ [0, 1] : f(x) ≥ 0E− = x ∈ [0, 1] : f(x) < 0;

thenE+, E− are measurable andµ(E+)+µ(E−) = 1.Write µ+ andµ− for µ(E+) andµ(E−). Also define

I+ =∫

E+

|f(x)| dx

I− =∫

E−

|f(x)| dx,

so that∫ 1

0|f(x)| dx = I+ + I−.

From the triangle inequality|a + b| ≥ ±(|a| − |b|), wehave the inequality∫∫

E+×E−

|f(x) + f(y)| dx dy

≥ ±∫∫

E+×E−

(|f(x)| − |f(y)|) dx dy

= ±(µ−I+ − µ+I−),

and likewise with+ and− switched. Adding these in-equalities together and allowing all possible choices ofthe signs, we get∫∫

(E+×E−)∪(E−×E+)

|f(x) + f(y)| dx dy

≥ max 0, 2(µ−I+ − µ+I−), 2(µ+I− − µ−I+) .

To this inequality, we add the equalities∫∫E+×E+

|f(x) + f(y)| dx dy = 2µ+I+∫∫E−×E−

|f(x) + f(y)| dx dy = 2µ−I−

−∫ 1

0

|f(x)| dx = −(µ+ + µ−)(I+ + I−)

to obtain∫ 1

0

∫ 1

0

|f(x) + f(y)| dx dy −∫ 1

0

|f(x)| dx

≥ max(µ+ − µ−)(I+ + I−) + 2µ−(I− − I+),(µ+ − µ−)(I+ − I−),(µ− − µ+)(I+ + I−) + 2µ+(I+ − I−).

Now simply note that for each of the possible compar-isons betweenµ+ andµ−, and betweenI+ andI−, oneof the three terms above is manifestly nonnegative. Thisyields the desired result.

Second solution: We will show at the end that it isenough to prove a discrete analogue: ifx1, . . . , xn arereal numbers, then

1n2

n∑i,j=1

|xi + xj | ≥1n

n∑i=1

|xi|.

6


In the meantime, we concentrate on this assertion.

Let f(x1, . . . , xn) denote the difference between thetwo sides. We induct on the number of nonzero valuesof |xi|. We leave for later the base case, where there isat most one such value. Suppose instead for now thatthere are two or more. Lets be the smallest, and sup-pose without loss of generality thatx1 = · · · = xa = s,xa+1 = · · · = xa+b = −s, and fori > a + b, eitherxi = 0 or |xi| > s. (One ofa, b might be zero.)

Now consider

f(

a terms︷︸︸︷t, · · · , t,

b terms︷︸︸︷−t, · · · ,−t, xa+b+1, · · · , xn)

as a function oft. It is piecewise linear nears; in fact,it is linear between 0 and the smallest nonzero valueamong|xa+b+1|, . . . , |xn| (which exists by hypothesis).Thus its minimum is achieved by one (or both) of thosetwo endpoints. In other words, we can reduce the num-ber of distinct nonzero absolute values among thexi

without increasingf . This yields the induction, pend-ing verification of the base case.

As for the base case, suppose thatx1 = · · · = xa =s > 0, xa+1 = · · · = xa+b = −s, andxa+b+1 = · · · =xn = 0. (Here one or even both ofa, b could be zero,though the latter case is trivial.) Then

f(x1, . . . , xn) =s

n2(2a2 + 2b2 + (a + b)(n− a− b))

− s

n(a + b)

=s

n2(a2 − 2ab + b2)

≥ 0.

This proves the base case of the induction, completingthe solution of the discrete analogue.

To deduce the original statement from the discrete ana-logue, approximate both integrals by equally-spacedRiemann sums and take limits. This works becausegiven a continuous function on a product of closed in-tervals, any sequence of Riemann sums with mesh sizetending to zero converges to the integral. (The domainis compact, so the function is uniformly continuous.Hence for anyε > 0 there is a cutoff below which anymesh size forces the discrepancy between the Riemannsum and the integral to be less thanε.)

Alternate derivation (based on a solution by Dan Bern-stein): from the discrete analogue, we have

∑1≤i<j≤n

|f(xi) + f(xj)| ≥n− 2

2

n∑i=1

|f(xi)|,

for all x1, . . . , xn ∈ [0, 1]. Integrating both sides as

(x1, . . . , xn) runs over[0, 1]n yields

n(n− 1)2

∫ 1

0

∫ 1

0

|f(x) + f(y)| dy dx

≥ n(n− 2)2

∫ 1

0

|f(x)| dx,

or∫ 1

0

∫ 1

0

|f(x) + f(y)| dy dx ≥ n− 2n− 1

∫ 1

0

|f(x)| dx.

Taking the limit asn → ∞ now yields the desired re-sult.

Third solution: (by David Savitt) We give an argumentwhich yields the following improved result. Letµp andµn be the measure of the setsx : f(x) > 0 andx : f(x) < 0 respectively, and letµ ≤ 1/2 bemin(µp, µn). Then∫ 1

0

∫ 1

0

|f(x) + f(y)| dx dy

≥ (1 + (1− 2µ)2)∫ 1

0

|f(x)| dx.

Note that the constant can be seen to be best possibleby considering a sequence of functions tending towardsthe step function which is1 on [0, µ] and−1 on (µ, 1].Suppose without loss of generality thatµ = µp. As inthe second solution, it suffices to prove a strengtheneddiscrete analogue, namely

1n2

∑i,j

|ai+aj | ≥

(1 +

(1− 2p

n

)2)(

1n

n∑i=1

|ai|

),

wherep ≤ n/2 is the number ofa1, . . . , an which arepositive. (We need only make sure to choose meshes sothatp/n → µ asn →∞.) An equivalent inequality is∑

1≤i<j≤n

|ai + aj | ≥(

n− 1− 2p +2p2

n

) n∑i=1

|ai|.

Write ri = |ai|, and assume without loss of gener-ality that ri ≥ ri+1 for each i. Then for i < j,|ai + aj | = ri + rj if ai andaj have the same sign,and isri − rj if they have opposite signs. The left-handside is therefore equal to

n∑i=1

(n− i)ri +n∑

j=1

rjCj ,

where

Cj = #i < j : sgn(ai) = sgn(aj)−#i < j : sgn(ai) 6= sgn(aj).

Consider the partial sumPk =∑k

j=1 Cj . If exactlypk

of a1, . . . , ak are positive, then this sum is equal to(pk

2

)+(

k − pk

2

)−[(

k

2

)−(

pk

2

)−(

k − pk

2

)],

7


which expands and simplifies to

−2pk(k − pk) +(

k

2

).

For k ≤ 2p even, this partial sum would be mini-mized with pk = k

2 , and would then equal−k2 ; for

k < 2p odd, this partial sum would be minimized withpk = k±1

2 , and would then equal−k−12 . Either way,

Pk ≥ −bk2 c. On the other hand, ifk > 2p, then

−2pk(k − pk) +(

k

2

)≥ −2p(k − p) +

(k

2

)sincepk is at mostp. DefineQk to be−bk

2 c if k ≤ 2p

and−2p(k−p)+(k2

)if k ≥ 2p, so thatPk ≥ Qk. Note

thatQ1 = 0.

Partial summation gives

n∑j=1

rjCj = rnPn +n∑

j=2

(rj−1 − rj)Pj−1

≥ rnQn +n∑

j=2

(rj−1 − rj)Qj−1

=n∑

j=2

rj(Qj −Qj−1)

= −r2 − r4 − · · · − r2p

+n∑

j=2p+1

(j − 1− 2p)rj .

It follows that∑1≤i<j≤n

|ai + aj | =n∑

i=1

(n− i)ri +n∑

j=1

rjCj

≥2p∑

i=1

(n− i− [i even])ri

+n∑

i=2p+1

(n− 1− 2p)ri

= (n− 1− 2p)n∑

i=1

ri+

2p∑i=1

(2p + 1− i− [i even])ri

≥ (n− 1− 2p)n∑

i=1

ri + p

2p∑i=1

ri

≥ (n− 1− 2p)n∑

i=1

ri + p2p

n

n∑i=1

ri ,

as desired. The next-to-last and last inequalities eachfollow from the monotonicity of theri’s, the former bypairing theith term with the(2p + 1− i)th.

Note: Compare the closely related Problem 6 from the2000 USA Mathematical Olympiad: prove that for anynonnegative real numbersa1, . . . , an, b1, . . . , bn, onehas

n∑i,j=1

minaiaj , bibj ≤n∑

i,j=1

minaibj , ajbi.

8




A1 Yes. Suppose otherwise. Then there would be anNsuch thatS(N) < 80% andS(N + 1) > 80%; that is,O’Keal’s free throw percentage is under80% at somepoint, and after one subsequent free throw (necessarilymade), her percentage is over80%. If she makesm ofher firstN free throws, thenm/N < 4/5 and (m +1)/(N + 1) > 4/5. This means that5m < 4n <5m + 1, which is impossible since then4n is an integerbetween the consecutive integers5m and5m + 1.

Remark: This same argument works for any fractionof the form(n − 1)/n for some integern > 1, but notfor any other real number between0 and1.

A2 First solution: (partly due to Ravi Vakil) Yes, it doesfollow. For i = 1, 2, let Pi, Qi, Ri be the vertices ofTi

opposide the sides of lengthai, bi, ci, respectively.

We first check the case wherea1 = a2 (or b1 = b2 orc1 = c2, by the same argument after relabeling). Imag-ine T2 as being drawn with the baseQ2R2 horizontaland the pointP2 above the lineQ2R2. We may thenpositionT1 so thatQ1 = Q2, R1 = R2, andP1 liesabove the lineQ1R1 = Q2R2. ThenP1 also lies insidethe region bounded by the circles throughP2 centeredat Q2 andR2. Since∠Q2 and∠R2 are acute, the partof this region above the lineQ2R2 lies within T2. Inparticular, the distance fromP1 to the lineQ2R2 is lessthan or equal to the distance fromP2 to the lineQ2R2;henceA1 ≤ A2.

To deduce the general case, put

r = maxa1/a2, b1/b2, c1/c2.

Let T3 be the triangle with sidesra2, rb2, rc2, whichhas arear2A2. Applying the special case toT1 andT3,we deduce thatA1 ≤ r2A2; sincer ≤ 1 by hypothesis,we haveA1 ≤ A2 as desired.

Remark: Another geometric argument in the casea1 =a2 is that since angles∠Q2 and∠R2 are acute, the per-pendicular toQ2R2 throughP2 separatesQ2 from R2.If A1 > A2, thenP1 lies above the parallel toQ2R2

throughP2; if then it lies on or to the left of the verticalline throughP2, we havec1 > c2 because the inequalityholds for both horizontal and vertical components (pos-sibly with equality for one, but not both). Similarly, ifP1 lies to the right of the vertical, thenb1 > b2.

Second solution: (attribution unknown) Retain nota-tion as in the first paragraph of the first solution. Sincethe angle measures in any triangle add up toπ, someangle ofT1 must have measure less than or equal to itscounterpart inT2. Without loss of generality assumethat∠P1 ≤ ∠P2. Since the latter is acute (becauseT2

is acute), we havesin∠P1 ≤ sin∠P2. By the Law ofSines,

A1 =12b1c1 sin∠P1 ≤

12b2c2 sin∠P2 = A2.

Remark: Many other solutions are possible; for in-stance, one uses Heron’s formula for the area of a tri-angle in terms of its side lengths.

A3 Define a sequencevn by vn = (n−1)(n−3) · · · (4)(2)if n is odd andvn = (n − 1)(n − 3) · · · (3)(1) if n iseven; it suffices to prove thatun = vn for all n ≥ 2.Now vn+3vn = (n + 2)(n)(n − 1)! andvn+2vn+1 =(n + 1)!, and sovn+3vn − vn+2vn+1 = n!. Since wecan check thatun = vn for n = 2, 3, 4, andun andvn

satisfy the same recurrence, it follows by induction thatun = vn for all n ≥ 2, as desired.

A4 It suffices to verify that

x1 · · ·xn

=1

2nn!

∑ei∈−1,1

(e1 · · · en)(e1x1 + · · ·+ enxn)n.

To check this, first note that the right side vanishes iden-tically for x1 = 0, because each term cancels the corre-sponding term withe1 flipped. Hence the right side, asa polynomial, is divisible byx1; similarly it is divisibleby x2, . . . , xn. Thus the right side is equal tox1 · · ·xn

times a scalar. (Another way to see this: the right side isclearly odd as a polynomial in each individual variable,but the only degreen monomial inx1, . . . , xn with thatproperty isx1 · · ·xn.) Since each summand contributes12n x1 · · ·xn to the sum, the scalar factor is 1 and we aredone.

Remark: Several variants on the above construction arepossible; for instance,

x1 · · ·xn

=1n!

∑ei∈0,1

(−1)n−e1−···−en(e1x1 + · · ·+ enxn)n

by the same argument as above.

Remark: These construction work over any field ofcharacteristic greater thann (at least forn > 1). On theother hand, no construction is possible over a field ofcharacteristicp ≤ n, since the coefficient ofx1 · · ·xn

in the expansion of(e1x1 + · · ·+enxn)n is zero for anyei.

Remark: Richard Stanley asks whether one can usefewer than2n terms, and what the smallest possiblenumber is.


2

A5 First solution: First recall that any graph withn ver-tices ande edges has at leastn − e connected com-ponents (add each edge one at a time, and note thatit reduces the number of components by at most 1).Now imagine the squares of the checkerboard as agraph, whose vertices are connected if the correspond-ing squares share a side and are the same color. LetAbe the number of edges in the graph, and letB be thenumber of 4-cycles (formed by monochromatic2 × 2squares). If we remove the bottom edge of each 4-cycle,the resulting graph has the same number of connectedcomponents as the original one; hence this number is atleast

mn−A + B.

By the linearity of expectation, the expected number ofconnected components is at least

mn− E(A) + E(B).

Moreover, we may computeE(A) by summing overthe individual pairs of adjacent squares, and we maycomputeE(B) by summing over the individual2 × 2squares. Thus

E(A) =12(m(n− 1) + (m− 1)n),

E(B) =18(m− 1)(n− 1),

and so the expected number of components is at least

mn− 12(m(n− 1) + (m− 1)n) +

18(m− 1)(n− 1)

=mn + 3m + 3n + 1

8>

mn

8.

Remark: A “dual” approach is to consider the graphwhose vertices are the corners of the squares of thecheckerboard, with two vertices joined if they are ad-jacent and the edge between then does not separate twosquares of the same color. In this approach, the 4-cyclesbecome isolated vertices, and the bound on componentsis replaced by a call to Euler’s formula relating the ver-tices, edges and faces of a planar figure. (One must becareful, however, to correctly handle faces which arenot simply connected.)

Second solution: (by Noam Elkies) Number thesquares of the checkerboard1, . . . ,mn by numberingthe first row from left to right, then the second row,and so on. We prove by induction oni that if we justconsider the figure formed by the firsti squares, itsexpected number of monochromatic components is atleasti/8. For i = 1, this is clear.

Suppose thei-th square does not abut the left edge orthe top row of the board. Then we may divide into threecases.

– With probability1/4, thei-th square is opposite incolor from the adjacent squares directly above andto the left of it. In this case adding thei-th squareadds one component.

– With probability1/8, the i-th square is the samein color as the adjacent squares directly above andto the left of it, but opposite in color from its diag-onal neighbor above and to the left. In this case,adding thei-th square either removes a componentor leaves the number unchanged.

– In all other cases, the number of components re-mains unchanged upon adding thei-th square.

Hence adding thei-th square increases the expectednumber of components by1/4− 1/8 = 1/8.

If the i-th square does abut the left edge of the board,the situation is even simpler: if thei-th square differs incolor from the square above it, one component is added,otherwise the number does not change. Hence addingthe i-th square increases the expected number of com-ponents by1/2; likewise if thei-th square abuts the topedge of the board. Thus the expected number of com-ponents is at leasti/8 by induction, as desired.

Remark: Some solvers attempted to consider addingone row at a time, rather than one square; this must behandled with great care, as it is possible that the num-ber of components can drop rather precipitously uponadding an entire row.

A6 By approximating each integral with a Riemann sum,we may reduce to proving the discrete analogue: forxij ∈ R for i, j = 1, . . . , n,

nn∑

i=1

n∑j=1

xij

2

+ nn∑

j=1

(n∑

i=1

xij

)2

≤

n∑i=1

n∑j=1

xij

2

+ n2n∑

i=1

n∑j=1

x2ij .

The difference between the right side and the left sideis

14

n∑i,j,k,l=1

(xij + xkl − xil − xkj)2,

which is evidently nonnegative. If you prefer not to dis-cretize, you may rewrite the original inequality as∫ 1

0

∫ 1

0

∫ 1

0

∫ 1

0

F (x, y, z, w)2 dx dy dz dw ≥ 0

for

F (x, y, z, w) = f(x, y) + f(z, w)− f(x, w)− f(z, y).


3

Remark: (by Po-Ning Chen) The discrete inequalitycan be arrived at more systematically by repeatedly ap-plying the following identity: for any reala1, . . . , an,

∑1≤i<j≤n

(xi − xj)2 = nn∑

i=1

x2i −

(n∑

i=1

xi

)2

.

Remark: (by David Savitt) The discrete inequality canalso be interpreted as follows. Forc, d ∈ 1, . . . , n−1andζn = e2πi/n, put

zc,d =∑i,j

ζci+djn xij .

Then the given inequality is equivalent to

n−1∑c,d=1

|zc,d|2 ≥ 0.

B1 Letk be an integer,0 ≤ k ≤ n−1. SinceP (r)/rk = 0,we have

cnrn−k + cn−1rn−k+1 + · · ·+ ck+1r

= −(ck + ck−1r−1 + · · ·+ c0r

−k).

Write r = p/q wherep andq are relatively prime. Thenthe left hand side of the above equation can be written asa fraction with denominatorqn−k, while the right handside is a fraction with denominatorpk. Sincep andqare relatively prime, both sides of the equation must bean integer, and the result follows.

Remark: If we write r = a/b in lowest terms, thenP (x) factors as(bx − a)Q(x), where the polynomialQ has integer coefficients because you can either do thelong division from the left and get denominators divis-ible only by primes dividingb, or do it from the rightand get denominators divisible only by primes dividinga. The numbers given in the problem are none otherthana times the coefficients ofQ. More generally, ifP (x) is divisible, as a polynomial over the rationals, bya polynomialR(x) with integer coefficients, thenP/Ralso has integer coefficients; this is known as “Gauss’slemma” and holds in any unique factorization domain.

B2 First solution: We have

(m + n)m+n >

(m + n

m

)mmnn

because the binomial expansion of(m + n)m+n in-cludes the term on the right as well as some others. Re-arranging this inequality yields the claim.

Remark: One can also interpret this argument com-binatorially. Suppose that we choosem + n times(with replacement) uniformly randomly from a set ofm + n balls, of whichm are red andn are blue. Thenthe probability of picking each ball exactly once is

(m + n)!/(m + n)m+n. On the other hand, ifp is theprobability of picking exactlym red balls, thenp < 1and the probability of picking each ball exactly once isp(mm/m!)(nn/n!).

Second solution:(by David Savitt) Define

Sk = i/k : i = 1, . . . , k

and rewrite the desired inequality as∏x∈Sm

x∏

y∈Sn

y >∏

z∈Sm+n

z.

To prove this, it suffices to check that if we sort themultiplicands on both sides into increasing order, thei-th term on the left side is greater than or equal to thei-thterm on the right side. (The equality is strict already fori = 1, so you do get a strict inequality above.)

Another way to say this is that for anyi, the number offactors on the left side which are less thani/(m + n) isless thani. But sincej/m < i/(m + n) is equivalentto j < im/(m + n), that number is⌈

im

m + n

⌉− 1 +

⌈in

m + n

⌉− 1

≤ im

m + n+

in

m + n− 1 = i− 1.

Third solution: Putf(x) = x(log(x+1)−log x); thenfor x > 0,

f ′(x) = log(1 + 1/x)− 1x + 1

f ′′(x) = − 1x(x + 1)2

.

Hencef ′′(x) < 0 for all x; sincef ′(x) → 0 asx →∞,we havef ′(x) > 0 for x > 0, sof is strictly increasing.

Put g(m) = m log m − log(m!); then g(m + 1) −g(m) = f(m), so g(m + 1) − g(m) increases withm. By induction,g(m + n) − g(m) increases withnfor any positive integern, so in particular

g(m + n)− g(m) > g(n)− g(1) + f(m)≥ g(n)

sinceg(1) = 0. Exponentiating yields the desired in-equality.

B3 The answer isa | a > 2. If a > 2, then the func-tion f(x) = 2a/(a − 2) has the desired property; bothperimeter and area ofR in this case are2a2/(a − 2).Now suppose thata ≤ 2, and letf(x) be a nonnegativecontinuous function on[0, a]. Let P = (x0, y0) be apoint on the graph off(x) with maximaly-coordinate;then the area ofR is at mostay0 since it lies below theline y = y0. On the other hand, the points(0, 0), (a, 0),andP divide the boundary ofR into three sections. The


4

length of the section between(0, 0) andP is at least thedistance between(0, 0) andP , which is at leasty0; thelength of the section betweenP and(a, 0) is similarlyat leasty0; and the length of the section between(0, 0)and(a, 0) is a. Sincea ≤ 2, we have2y0 + a > ay0

and hence the perimeter ofR is strictly greater than thearea ofR.

B4 First solution: Identify thexy-plane with the complexplaneC, so thatPk is the real numberk. If z is sent toz′ by a counterclockwise rotation byθ aboutPk, thenz′ − k = eiθ(z − k); hence the rotationRk sendsz toζz + k(1 − ζ), whereζ = e2πi/n. It follows thatR1

followed byR2 sendsz to ζ(ζz+(1−ζ))+2(1−ζ) =ζ2z + (1 − ζ)(ζ + 2), and so forth; an easy inductionshows thatR sendsz to

ζnz + (1− ζ)(ζn−1 + 2ζn−2 + · · ·+ (n− 1)ζ + n).

Expanding the product(1− ζ)(ζn−1 + 2ζn−2 + · · ·+(n − 1)ζ + n) yields−ζn − ζn−1 − · · · − ζ + n =n. ThusR sendsz to z + n; in cartesian coordinates,R(x, y) = (x + n, y).

Second solution:(by Andy Lutomirski, via Ravi Vakil)Imagine a regularn-gon of side length 1 placed with itstop edge on thex-axis and the left endpoint of that edgeat the origin. Then the rotations correspond to rollingthis n-gon along thex-axis; after then rotations, itclearly ends up in its original rotation and translatednunits to the right. Hence the whole plane must do so aswell.

Third solution: (attribution unknown) Viewing eachRk as a function of a complex numberz as in the firstsolution, the functionRn Rn−1 · · · R1(z) is linearin z with slopeζn = 1. It thus equalsz + T for someT ∈ C. Sincef1(1) = 1, we can write1 + T = Rn · · · R2(1). However, we also have

Rn · · · R2(1) = Rn−1 R1(0) + 1

by the symmetry in how theRi are defined. Hence

Rn(1 + T ) = Rn R1(0) + Rn(1) = T + Rn(1);

that is,Rn(T ) = T . HenceT = n, as desired.

B5 First solution: By taking logarithms, we seethat the desired limit isexp(L), where L =limx→1−

∑∞n=0 xn

(ln(1 + xn+1)− ln(1 + xn)

).

Now

N∑n=0

xn(ln(1 + xn+1)− ln(1 + xn)

)= 1/x

N∑n=0

xn+1 ln(1 + xn+1)−N∑

n=0

xn ln(1 + xn)

= xN ln(1 + xN+1)− ln 2 + (1/x− 1)N∑

n=1

xn ln(1 + xn);

sincelimN→∞(xN ln(1 + xN+1)) = 0 for 0 < x < 1,we conclude thatL = − ln 2 + limx→1− f(x), where

f(x) = (1/x− 1)∞∑

n=1

xn ln(1 + xn)

= (1/x− 1)∞∑

n=1

∞∑m=1

(−1)m+1xn+mn/m.

This final double sum converges absolutely when0 <x < 1, since

∞∑n=1

∞∑m=1

xn+mn/m =∞∑

n=1

xn(− ln(1− xn))

<∞∑

n=1

xn(− ln(1− x)),

which converges. (Note that− ln(1− x) and− ln(1−xn) are positive.) Hence we may interchange the sum-mations inf(x) to obtain

f(x) = (1/x− 1)∞∑

m=1

∞∑n=1

(−1)m+1x(m+1)n

m

= (1/x− 1)∞∑

m=1

(−1)m+1

m

(xm(1− x)1− xm+1

).

This last sum converges absolutely uniformly inx, soit is legitimate to take limits term by term. Sincelimx→1−

xm1−x1−xm+1 = 1

m+1 for fixedm, we have

limx→1−

f(x) =∞∑

m=1

(−1)m+1

m(m + 1)

=∞∑

m=1

(−1)m+1

(1m− 1

m + 1

)

= 2

( ∞∑m=1

(−1)m+1

m

)− 1

= 2 ln 2− 1,

and henceL = ln 2− 1 and the desired limit is2/e.

Remark: Note that the last series is not absolutely con-vergent, so the recombination must be done without re-arranging terms.

Second solution:(by Greg Price, via Tony Zhang andAnders Kaseorg) Puttn(x) = ln(1 + xn); we can thenwrite xn = exp(tn(x))− 1, and

L = limx→1−

∞∑n=0

(tn(x)− tn+1(x))(1− exp(tn(x))).

The expression on the right is a Riemann sum approxi-mating the integral

∫ ln 2

0(1−et) dt, over the subdivision

of [0, ln(2)) given by thetn(x). As x → 1−, the max-imum difference between consecutivetn(x) tends to 0,so the Riemann sum tends to the value of the integral.HenceL =

∫ ln 2

0(1− et) dt = ln 2− 1, as desired.


5

B6 First solution: (based on a solution of Dan Bernstein)Note that for anyb, the condition thatb /∈ B alreadyforceslim sup N(x)/x to be at most 1/2: pair off2mb+n with (2m + 1)b + n for n = 1, . . . , b, and note thatat most one member of each pair may belong toA. Theidea of the proof is to do something similar with pairsreplaced by larger clumps, using long runs of excludedelements ofB.

Suppose we have positive integersb0 = 1, b1, . . . , bn

with the following properties:

(a) Fori = 1, . . . , n, ci = bi/(2bi−1) is an integer.

(b) Forei ∈ −1, 0, 1, |e1b1 + · · ·+ enbn| /∈ B.

Each nonnegative integera has a unique “base expan-sion”

a = a0b0 + · · ·+ an−1bn−1 + mbn (0 ≤ ai < 2ci);

if two integers have expansions with the same valueof m, and values ofai differing by at most 1 fori = 0, . . . , n − 1, then their difference is not inB, soat most one of them lies inA. In particular, for anydi ∈ 0, . . . , ci − 1, anym0 ∈ 0, 2c0 − 1 and anymn, the set

m0b0 + (2d1 + e1)b0 + · · ·+ (2dn−1 + en−1)bn−1 + (2mn + en)bn,

where eachei runs over0, 1, contains at most oneelement ofA; consequently,lim supN(x)/x ≤ 1/2n.

We now produce suchbi recursively, starting withb0 =1 (and both (a) and (b) holding vacuously). Givenb0, . . . , bn satisfying (a) and (b), note thatb0 + · · · +bn−1 < bn by induction onn. By the hypothesesof the problem, we can find a setSn of 6bn consec-utive integers, none of which belongs toB. Let bn+1

be the second-smallest multiple of2bn in Sn; thenbn+1 + x ∈ Sn for −2bn ≤ x ≤ 0 clearly, and alsofor 0 ≤ x ≤ 2bn because there are most4bn − 1 ele-ments ofSn precedingbn+1. In particular, the analogue

of (b) with n replaced byn + 1 holds foren+1 6= 0; ofcourse it holds foren+1 = 0 because (b) was alreadyknown. Since the analogue of (a) holds by construction,we have completed this step of the construction and therecursion may continue.

Since we can constructb0, . . . , bn satisfying (a) and (b)for anyn, we havelim sup N(x)/x ≤ 1/2n for anyn,yielding limN(x)/x = 0 as desired.

Second solution: (by Paul Pollack) LetS be the setof possible values oflim supN(x)/x; sinceS ⊆ [0, 1]is bounded, it has a least upper boundL. Suppose byway of contradiction thatL > 0; we can then chooseA,B satisfying the conditions of the problem such thatlim sup N(x)/x > 3L/4.

To begin with, we can certainly find some positive inte-germ /∈ B, so thatA is disjoint fromA+m = a+m :a ∈ A. PutA′ = A∪(A+m) and letN ′(x) be the sizeofA′∩1, . . . , x; thenlim sup N ′(x)/x = 3L/2 > L,soA′ cannot obey the conditions of the problem state-ment. That is, if we letB′ be the set of positive integersthat occur as differences between elements ofA′, thenthere exists an integern such that among anyn consec-utive integers, at least one lies inB′. But

B′ ⊆ b + em : b ∈ B, e ∈ −1, 0, 1,

so among anyn + 2m consecutive integers, at least onelies inB. This contradicts the condition of the problemstatement.

We conclude that it is impossible to haveL > 0, soL = 0 andlim N(x)/x = 0 as desired.

Remark: A hybrid between these two arguments isto note that if we can producec1, . . . , cn such that|ci−cj | /∈ B for i, j = 1, . . . , n, then the translatesA+c1, . . . ,A + cn are disjoint and solim sup N(x)/x ≤1/n. Givenc1 ≤ · · · ≤ cn as above, we can then choosecn+1 to be the largest element of a run ofcn + 1 con-secutive integers, none of which lie inB.




A1 We proceed by induction, with base case1 = 2030.Suppose all integers less thann− 1 can be represented.If n is even, then we can take a representation ofn/2and multiply each term by 2 to obtain a representationof n. If n is odd, putm = blog3 nc, so that3m ≤ n <3m+1. If 3m = n, we are done. Otherwise, choose arepresentation(n−3m)/2 = s1+· · ·+sk in the desiredform. Then

n = 3m + 2s1 + · · ·+ 2sk,

and clearly none of the2si divide each other or3m.Moreover, since2si ≤ n − 3m < 3m+1 − 3m, wehavesi < 3m, so 3m cannot divide2si either. Thusn has a representation of the desired form in all cases,completing the induction.

Remarks: This problem is originally due to Paul Erdos.Note that the representations need not be unique: forinstance,

11 = 2 + 9 = 3 + 8.

A2 We will assumen ≥ 2 hereafter, since the answer is 0for n = 1.

First solution: We show that the set of rook tours from(1, 1) to (n, 1) is in bijection with the set of subsets of1, 2, ..., n that includen and contain an even numberof elements in total. Since the latter set evidently con-tains2n−2 elements, so does the former.

We now construct the bijection. Given a rook tourPfrom (1, 1) to (n, 1), let S = S(P ) denote the set ofall i ∈ 1, 2, . . . , n for which there is either a directededge from(i, 1) to (i, 2) or from (i, 3) to (i, 2). It isclear that this setS includesn and must contain aneven number of elements. Conversely, given a subsetS = a1, a2, . . . , a2r = n ⊂ 1, 2, . . . , n of thistype witha1 < a2 < · · · < a2r, we notice that there isa unique pathP containing(ai, 2 + (−1)i), (a1, 2) fori = 1, 2, . . . , 2r. This establishes the desired bijection.

Second solution:Let An denote the set of rook toursbeginning at(1, 1) and ending at(n, 1), and letBn de-note the set of rook tours beginning at(1, 1) and endingat (n, 3).

For n ≥ 2, we construct a bijection betweenAn andAn−1∪Bn−1. Any pathP in An contains either the linesegmentP1 between(n − 1, 1) and(n, 1), or the linesegmentP2 between(n, 2) and (n, 1). In the formercase,P must also contain the subpathP ′

1 which joins(n − 1, 3), (n, 3), (n, 2), and(n − 1, 2) consecutively;then deletingP1 and P ′

1 from P and adding the line

segment joining(n−1, 3) to (n−1, 2) results in a pathin An−1. (This construction is reversible, lengtheningany path inAn−1 to a path inAn.) In the latter case,Pcontains the subpathP ′

2 which joins(n − 1, 3), (n, 3),(n, 2), (n, 1) consecutively; deletingP ′

2 results in a pathin Bn−1, and this construction is also reversible. Thedesired bijection follows.

Similarly, there is a bijection betweenBn andAn−1 ∪Bn−1 for n ≥ 2. It follows by induction that forn ≥ 2,|An| = |Bn| = 2n−2(|A1| + |B1|). But |A1| = 0 and|B1| = 1, and hence the desired answer is|An| = 2n−2.

Remarks: Other bijective arguments are possible: forinstance, Noam Elkies points out that each elementof An ∪ Bn contains a different one of the possi-ble sets of segments of the form(i, 2), (i + 1, 2) fori = 1, . . . , n − 1. Richard Stanley provides the refer-ence: K.L. Collins and L.B. Krompart, The number ofHamiltonian paths in a rectangular grid,Discrete Math.169 (1997), 29–38. This problem is Theorem 1 of thatpaper; the cases of4 × n and 5 × n grids are alsotreated. The paper can also be found online at the URLkcollins.web.wesleyan.edu/vita.htm .

A3 Note that it is implicit in the problem thatp is noncon-stant, one may take any branch of the square root, andthatz = 0 should be ignored.

First solution: Write p(z) = c∏n

j=1(z − rj), so that

g′(z)g(z)

=12z

n∑j=1

z + rj

z − rj.

Now if z 6= rj for all j,then

z + rj

z − rj=

(z + rj)(z − rj)|z − rj |2

=|z|2 − 1 + 2Im(zrj)

|z − rj |2,

and so

Rezg′(z)g(z)

=|z|2 − 1

2

∑j

1|z − rj |2

.

Since the quantity in parentheses is positive,g′(z)/g(z)can be0 only if |z| = 1. If on the other handz = rj forsomej, then|z| = 1 anyway.

Second solution:Write p(z) = c∏n

j=1(z−rj), so that

g′(z)g(z)

=n∑

j=1

(1

z − rj− 1

2z

).


2

We first check thatg′(z) 6= 0 wheneverz is real andz > 1. In this case, forrj = eiθj , we havez − rj =(z − cos(θj)) + sin(θj)i, so the real part of 1

z−rj− 1

2z

is

z − cos(θj)z2 − 2z cos(θj) + 1

− 12z

=z2 − 1

2z(z2 − 2z cos(θj) + 1)> 0.

Henceg′(z)/g(z) has positive real part, sog′(z)/g(z)and henceg(z) are nonzero.

Applying the same argument after replacingp(z) byp(eiθz), we deduce thatg′ cannot have any roots out-side the unit circle. Applying the same argument afterreplacingp(z) by znp(1/z), we also deduce thatg′ can-not have any roots inside the unit circle. Hence all rootsof g′ have absolute value 1, as desired.

Third solution: Write p(z) = c∏n

j=1(z − rj) and putrj = e2iθj . Note thatg(e2iθ) is equal to a nonzeroconstant times

h(θ) =n∏

j=1

ei(θ+θj) − e−i(θ+θj)

2i

=n∏

j=1

sin(θ + θj).

Sinceh has at least2n roots (counting multiplicity) inthe interval[0, 2π), h′ does also by repeated applica-tion of Rolle’s theorem. Sinceg′(e2iθ) = 2ie2iθh′(θ),g′(z2) has at least2n roots on the unit circle. Sinceg′(z2) is equal toz−n−1 times a polynomial of degree2n, g′(z2) has all roots on the unit circle, as then doesg′(z).

Remarks: The second solution imitates the proof ofthe Gauss-Lucas theorem: the roots of the derivative ofa complex polynomial lie in the convex hull of the rootsof the original polynomial. The second solution is closeto problem B3 from the 2000 Putnam. A hybrid be-tween the first and third solutions is to check that on theunit circle, Re(zg′(z)/g(z)) = 0 while between anytwo roots ofp, Im(zg′(z)/g(z)) runs from+∞ to−∞and so must have a zero crossing. (This only workswhenp has distinct roots, but the general case followsby the continuity of the roots of a polynomial as func-tions of the coefficients.) One can also construct a solu-tion using Rouche’s theorem.

A4 First solution: Choose a set ofa rowsr1, . . . , ra con-taining ana× b submatrix whose entries are all 1. Thenfor i, j ∈ 1, . . . , a, we haveri · rj = n if i = j and 0otherwise. Hence

a∑i,j=1

ri · rj = an.

On the other hand, the term on the left is the dot productof r1+ · · ·+ra with itself, i.e., its squared length. Sincethis vector hasa in each of its firstb coordinates, thedot product is at leasta2b. Hencean ≥ a2b, whencen ≥ ab as desired.

Second solution:(by Richard Stanley) Suppose with-out loss of generality that thea× b submatrix occupiesthe firsta rows and the firstb columns. LetM be thesubmatrix occupying the firsta rows and the lastn − bcolumns. Then the hypothesis implies that the matrixMMT hasn− b’s on the main diagonal and−b’s else-where. Hence the column vectorv of lengtha consist-ing of all 1’s satisfiesMMT v = (n−ab)v, son−ab isan eigenvalue ofMMT . But MMT is semidefinite, soits eigenvalues are all nonnegative real numbers. Hencen− ab ≥ 0.

Remarks: A matrix as in the problem is calleda Hadamard matrix, because it meets the equalitycondition of Hadamard’s inequality: anyn × n matrixwith ±1 entries has absolute determinant at mostnn/2,with equality if and only if the rows are mutuallyorthogonal (from the interpretation of the determinantas the volume of a paralellepiped whose edges areparallel to the row vectors). Note that this impliesthat the columns are also mutually orthogonal. Ageneralization of this problem, with a similar proof,is known asLindsey’s lemma: the sum of the entriesin any a × b submatrix of a Hadamard matrix is atmost

√abn. Stanley notes that Ryser (1981) asked

for the smallest size of a Hadamard matrix containingan r × s submatrix of all 1’s, and refers to the URLwww3.interscience.wiley.com/cgi-bin/abstract/110550861/ABSTRACT for moreinformation.

A5 First solution: We make the substitutionx = tan θ,rewriting the desired integral as∫ π/4

0

log(tan(θ) + 1) dθ.

Write

log(tan(θ) + 1)= log(sin(θ) + cos(θ))− log(cos(θ))

and then note thatsin(θ) + cos(θ) =√

2 cos(π/4− θ).We may thus rewrite the integrand as

12

log(2) + log(cos(π/4− θ))− log(cos(θ)).

But over the interval [0, π/4], the integrals oflog(cos(θ)) and log(cos(π/4 − θ)) are equal, so theircontributions cancel out. The desired integral is thenjust the integral of12 log(2) over the interval[0, π/4],which isπ log(2)/8.

Second solution: (by Roger Nelsen) LetI denote thedesired integral. We make the substitutionx = (1 −


3

u)/(1 + u) to obtain

I =∫ 1

0

(1 + u)2 log(2/(1 + u))2(1 + u2)

2 du

(1 + u)2

=∫ 1

0

log(2)− log(1 + u)1 + u2

du

= log(2)∫ 1

0

du

1 + u2− I,

yielding

I =12

log(2)∫ 1

0

du

1 + u2=

π log(2)8

.

Third solution: (attributed to Steven Sivek) Define thefunction

f(t) =∫ 1

0

log(xt + 1)x2 + 1

dx

so thatf(0) = 0 and the desired integral isf(1). Thenby differentiation under the integral,

f ′(t) =∫ 1

0

x

(xt + 1)(x2 + 1)dx.

By partial fractions, we obtain

f ′(t) =2t arctan(x)− 2 log(tx + 1) + log(x2 + 1)

2(t2 + 1)

∣∣∣∣x=1

x=0

=πt + 2 log(2)− 4 log(t + 1)

4(t2 + 1),

whence

f(t) =log(2) arctan(t)

2+

π log(t2 + 1)8

−∫ t

0

log(t + 1)t2 + 1

dt

and hence

f(1) =π log(2)

4−∫ 1

0

log(t + 1)t2 + 1

dt.

But the integral on the right is again the desired integralf(1), so we may move it to the left to obtain

2f(1) =π log(2)

4

and hencef(1) = π log(2)/8 as desired.

Fourth solution: (by David Rusin) We have∫ 1

0

log(x + 1)x2 + 1

dx =∫ 1

0

( ∞∑n=1

(−1)n−1xn

n(x2 + 1)

)dx.

We next justify moving the sum through the integralsign. Note that

∞∑n=1

∫ 1

0

(−1)n−1xn dx

n(x2 + 1)

is an alternating series whose terms strictly decrease tozero, so it converges. Moreover, its partial sums alter-nately bound the previous integral above and below, sothe sum of the series coincides with the integral.

Put

Jn =∫ 1

0

xn dx

x2 + 1;

thenJ0 = arctan(1) = π4 andJ1 = 1

2 log(2). More-over,

Jn + Jn+2 =∫ 1

0

xn dx =1

n + 1.

Write

Am =m∑

i=1

(−1)i−1

2i− 1

Bm =m∑

i=1

(−1)i−1

2i;

then

J2n = (−1)n(J0 −An)J2n+1 = (−1)n(J1 −Bn).

Now the2N -th partial sum of our series equals

N∑n=1

J2n−1

2n− 1− J2n

2n

=N∑

n=1

(−1)n−1

2n− 1(J1 −Bn−1)−

(−1)n

2n(J0 −An)

= AN (J1 −BN−1) + BN (J0 −AN ) + ANBN .

As N → ∞, AN → J0 andBN → J1, so the sumtends toJ0J1 = π log(2)/8.

Remarks: The first two solutions are related by the factthat if x = tan(θ), then1−x/(1+x) = tan(π/4−θ).The strategy of the third solution (introducing a parame-ter then differentiating it) was a favorite of physics No-belist (and Putnam Fellow) Richard Feynman. NoamElkies notes that this integral is number 2.491#8 inGradshteyn and Ryzhik,Table of integrals, series, andproducts. The Mathematicacomputer algebra system(version 5.2) successfully computes this integral, but wedo not know how.

A6 First solution: The angle at a vertexP is acute if andonly if all of the other points lie on an open semicir-cle. We first deduce from this that if there are anytwo acute angles at all, they must occur consecutively.Suppose the contrary; label the verticesQ1, . . . , Qn incounterclockwise order (starting anywhere), and sup-pose that the angles atQ1 andQi are acute for somei with 3 ≤ i ≤ n− 1. Then the open semicircle starting


4

at Q2 and proceeding counterclockwise must containall of Q3, . . . , Qn, while the open semicircle startingat Qi and proceeding counterclockwise must containQi+1, . . . , Qn, Q1, . . . , Qi−1. Thus two open semicir-cles cover the entire circle, contradiction.

It follows that if the polygon has at least one acute an-gle, then it has either one acute angle or two acute an-gles occurring consecutively. In particular, there is aunique pair of consecutive verticesQ1, Q2 in counter-clockwise order for which∠Q2 is acute and∠Q1 is notacute. Then the remaining points all lie in the arc fromthe antipode ofQ1 to Q1, butQ2 cannot lie in the arc,and the remaining points cannot all lie in the arc fromthe antipode ofQ1 to the antipode ofQ2. Given thechoice ofQ1, Q2, let x be the measure of the counter-clockwise arc fromQ1 to Q2; then the probability thatthe other points fall into position is2−n+2 − xn−2 ifx ≤ 1/2 and 0 otherwise.

Hence the probability that the polygon has at least oneacute angle with agivenchoice of which two points willact asQ1 andQ2 is∫ 1/2

0

(2−n+2 − xn−2) dx =n− 2n− 1

2−n+1.

Since there aren(n − 1) choices for which two pointsact asQ1 andQ2, the probability of at least one acuteangle isn(n− 2)2−n+1.

Second solution:(by Calvin Lin) As in the first solu-tion, we may compute the probability that for a particu-lar one of the pointsQ1, the angle atQ1 is not acute butthe following angle is, and then multiply byn. Imaginepicking the points by first choosingQ1, then pickingn − 1 pairs of antipodal points and then picking onemember of each pair. LetR2, . . . , Rn be the points ofthe pairs which lie in the semicircle, taken in order awayfrom Q1, and letS2, . . . , Sn be the antipodes of these.Then to get the desired situation, we must choose fromthe pairs to end up with all but one of theSi, and wecannot takeRn and the otherSi or else∠Q1 will beacute. That gives us(n− 2) good choices out of2n−1;since we could have chosenQ1 to be any of then points,the probability is againn(n− 2)2−n+1.

B1 TakeP (x, y) = (y − 2x)(y − 2x − 1). To see thatthis works, first note that ifm = bac, then2m is aninteger less than or equal to2a, so 2m ≤ b2ac. Onthe other hand,m + 1 is an integer strictly greater thana, so2m + 2 is an integer strictly greater than2a, sob2ac ≤ 2m + 1.

B2 By the arithmetic-harmonic mean inequality or theCauchy-Schwarz inequality,

(k1 + · · ·+ kn)(

1k1

+ · · ·+ 1kn

)≥ n2.

We must thus have5n − 4 ≥ n2, son ≤ 4. Withoutloss of generality, we may suppose thatk1 ≤ · · · ≤ kn.

If n = 1, we must havek1 = 1, which works. Note thathereafter we cannot havek1 = 1.

If n = 2, we have(k1, k2) ∈ (2, 4), (3, 3), neither ofwhich work.

If n = 3, we have k1 + k2 + k3 = 11,so 2 ≤ k1 ≤ 3. Hence (k1, k2, k3) ∈(2, 2, 7), (2, 3, 6), (2, 4, 5), (3, 3, 5), (3, 4, 4), andonly (2, 3, 6) works.

If n = 4, we must have equality in the AM-HM inequal-ity, which only happens whenk1 = k2 = k3 = k4 = 4.

Hence the solutions aren = 1 andk1 = 1, n = 3 and(k1, k2, k3) is a permutation of(2, 3, 6), andn = 4 and(k1, k2, k3, k4) = (4, 4, 4, 4).Remark: In the casesn = 2, 3, Greg Kuperberg sug-gests the alternate approach of enumerating the solu-tions of1/k1+· · ·+1/kn = 1 with k1 ≤ · · · ≤ kn. Thisis easily done by proceeding in lexicographic order: oneobtains(2, 2) for n = 2, and(2, 3, 6), (2, 4, 4), (3, 3, 3)for n = 3, and only(2, 3, 6) contributes to the final an-swer.

B3 First solution: The functions are preciselyf(x) = cxd

for c, d > 0 arbitrary except that we must takec = 1in cased = 1. To see that these work, note thatf ′(a/x) = dc(a/x)d−1 andx/f(x) = 1/(cxd−1), sothe given equation holds if and only ifdc2ad−1 = 1.If d 6= 1, we may solve fora no matter whatc is; ifd = 1, we must havec = 1. (Thanks to Brad Rodgersfor pointing out thed = 1 restriction.)

To check that these are all solutions, putb = log(a) andy = log(a/x); rewrite the given equation as

f(eb−y)f ′(ey) = eb−y.

Put

g(y) = log f(ey);

then the given equation rewrites as

g(b− y) + log g′(y) + g(y)− y = b− y,

or

log g′(y) = b− g(y)− g(b− y).

By the symmetry of the right side, we haveg′(b− y) =g′(y). Hence the functiong(y) + g(b − y) has zeroderivative and so is constant, as then isg′(y). Fromthis we deduce thatf(x) = cxd for somec, d, bothnecessarily positive sincef ′(x) > 0 for all x.

Second solution:(suggested by several people) Substi-tutea/x for x in the given equation:

f ′(x) =a

xf(a/x).

Differentiate:

f ′′(x) = − a

x2f(a/x)+

a2f ′(a/x)x3f(a/x)2

.


5

Now substitute to eliminate evaluations ata/x:

f ′′(x) = −f ′(x)x

+f ′(x)2

f(x).

Clear denominators:

xf(x)f ′′(x) + f(x)f ′(x) = xf ′(x)2.

Divide through byf(x)2 and rearrange:

0 =f ′(x)f(x)

+xf ′′(x)f(x)

− xf ′(x)2

f(x)2.

The right side is the derivative ofxf ′(x)/f(x), so thatquantity is constant. That is, for somed,

f ′(x)f(x)

=d

x.

Integrating yieldsf(x) = cxd, as desired.

B4 First solution: Definef(m,n, k) as the number ofn-tuples(x1, x2, . . . , xn) of integers such that|x1|+· · ·+|xn| ≤ m and exactlyk of x1, . . . , xn are nonzero. Tochoose such a tuple, we may choose thek nonzero posi-tions, the signs of thosek numbers, and then an orderedk-tuple of positive integers with sum≤ m. There are(nk

)options for the first choice, and2k for the second.

As for the third, we have(mk

)options by a “stars and

bars” argument: depict thek-tuple by drawing a num-ber of stars for each term, separated by bars, and addingstars at the end to get a total ofm stars. Then each tu-ple corresponds to placingk bars, each in a differentposition behind one of them fixed stars.

We conclude that

f(m,n, k) = 2k

(m

k

)(n

k

)= f(n, m, k);

summing overk givesf(m,n) = f(n, m). (One mayalso extract easily a bijective interpretation of the equal-ity.)

Second solution:(by Greg Kuperberg) It will be con-venient to extend the definition off(m,n) to m,n ≥ 0,in which case we havef(0,m) = f(n, 0) = 1.

Let Sm,n be the set ofn-tuples(x1, . . . , xn) of inte-gers such that|x1| + · · · + |xn| ≤ m. Then elementsof Sm,n can be classified into three types. Tuples with|x1| + · · · + |xn| < m also belong toSm−1,n. Tupleswith |x1| + · · · + |xn| = m andxn ≥ 0 correspondto elements ofSm,n−1 by droppingxn. Tuples with|x1| + · · · + |xn| = m andxn < 0 correspond to ele-ments ofSm−1,n−1 by droppingxn. It follows that

f(m,n)= f(m− 1, n) + f(m,n− 1) + f(m− 1, n− 1),

so f satisfies a symmetric recurrence with symmetricboundary conditionsf(0,m) = f(n, 0) = 1. Hencefis symmetric.

Third solution: (by Greg Martin) As in the second so-lution, it is convenient to allowf(m, 0) = f(0, n) = 1.Define the generating function

G(x, y) =∞∑

m=0

∞∑n=0

f(m,n)xmyn.

As equalities of formal power series (or convergent se-ries on, say, the region|x|, |y| < 1

3 ), we have

G(x, y) =∑m≥0

∑n≥0

xmyn∑

k1, ..., kn∈Z|k1|+···+|kn|≤m

1

=∑n≥0

yn∑

k1, ..., kn∈Z

∑m≥|k1|+···+|kn|

xm

=∑n≥0

yn∑

k1, ..., kn∈Z

x|k1|+···+|kn|

1− x

=1

1− x

∑n≥0

yn

(∑k∈Z

x|k|)n

=1

1− x

∑n≥0

yn

(1 + x

1− x

)n

=1

1− x· 11− y(1 + x)/(1− x)

=1

1− x− y − xy.

SinceG(x, y) = G(y, x), it follows that f(m,n) =f(n, m) for all m,n ≥ 0.

B5 First solution: Put Q = x21 + · · · + x2

n. SinceQ ishomogeneous,P is divisible byQ if and only if each ofthe homogeneous components ofP is divisible byQ. Itis thus sufficient to solve the problem in caseP itself ishomogeneous, say of degreed.

Suppose that we have a factorizationP = QmR forsomem > 0, whereR is homogeneous of degreed andnot divisible byQ; note that the homogeneity impliesthat

n∑i=1

xi∂R

∂xi= dR.

Write∇2 as shorthand for∂2

∂x21

+ · · ·+ ∂2

∂x2n

; then

0 = ∇2P

= 2mnQm−1R + Qm∇2R + 2n∑

i=1

2mxiQm−1 ∂R

∂xi

= Qm∇2R + (2mn + 4md)Qm−1R.

Sincem > 0, this forcesR to be divisible byQ, con-tradiction.

Second solution:(by Noam Elkies) Retain notation asin the first solution. LetPd be the set of homogeneous


6

polynomials of degreed, and letHd be the subset ofPd of polynomials killed by∇2, which has dimension≥ dim(Pd) − dim(Pd−2); the given problem amountsto showing that this inequality is actually an equality.

Consider the operatorQ∇2 (i.e., apply∇2 then multi-ply by Q) on Pd; its zero eigenspace is preciselyHd.By the calculation from the first solution, ifR ∈ Pd,then

∇2(QR)−Q∇2R = (2n + 4d)R.

Consequently,QjHd−2j is contained in the eigenspaceof Q∇2 onPd of eigenvalue

(2n + 4(d− 2j)) + · · ·+ (2n + 4(d− 2)).

In particular, theQjHd−2j lie in distinct eigenspaces,so are linearly independent withinPd. But by dimen-sion counting, their total dimension is at least that ofPd.Hence they exhaustPd, and the zero eigenspace cannothave dimension greater thandim(Pd)− dim(Pd−2), asdesired.

Third solution: (by Richard Stanley) Writex =(x1, . . . , xn) and∇ = ( ∂

∂x1, . . . , ∂

∂xn). Suppose that

P (x) = Q(x)(x21 + · · ·+ x2

n). Then

P (∇)P (x) = Q(∇)(∇2)P (x) = 0.

On the other hand, ifP (x) =∑

α cαxα (whereα =(α1, . . . , αn) andxα = xα1

1 · · ·xαnn ), then the constant

term ofP (∇)P (x) is seen to be∑

α c2α. Hencecα = 0

for all α.

Remarks: The first two solutions apply directly overany field of characteristic zero. (The result fails in char-acteristicp > 0 because we may takeP = (x2

1 +· · · + x2

n)p = x2p1 + · · · + x2p

n .) The third solutioncan be extended to complex coefficients by replacingP (∇) by its complex conjugate, and again the resultmay be deduced for any field of characteristic zero.Stanley also suggests Section 5 of the arXiv e-printmath.CO/0502363 for some algebraic backgroundfor this problem.

B6 First solution: Let I be the identity matrix, and letJx be the matrix withx’s on the diagonal and 1’s else-where. Note thatJx−(x−1)I, being the all 1’s matrix,has rank 1 and tracen, so hasn − 1 eigenvalues equalto 0 and one equal ton. HenceJx hasn−1 eigenvaluesequal tox− 1 and one equal tox + n− 1, implying

det Jx = (x + n− 1)(x− 1)n−1.

On the other hand, we may expand the determinant as asum indexed by permutations, in which case we get

det Jx =∑

π∈Sn

sgn(π)xν(π).

Integrating both sides from 0 to 1 (and substitutingy =1− x) yields

∑π∈Sn

sgn(π)ν(π) + 1

=∫ 1

0

(x + n− 1)(x− 1)n−1 dx

=∫ 1

0

(−1)n+1(n− y)yn−1 dy

= (−1)n+1 n

n + 1,

as desired.

Second solution: We start by recalling a form of theprinciple of inclusion-exclusion: iff is a function onthe power set of1, . . . , n, then

f(S) =∑T⊇S

(−1)|T |−|S|∑U⊇T

f(U).

In this case we takef(S) to be the sum ofσ(π) over allpermutationsπ whose fixed points are exactlyS. Then∑

U⊇T f(U) = 1 if |T | ≥ n−1 and 0 otherwise (sincea permutation group on 2 or more symbols has as manyeven and odd permutations), so

f(S) = (−1)n−|S|(1− n + |S|).

The desired sum can thus be written, by grouping overfixed point sets, as

n∑i=0

(n

i

)(−1)n−i 1− n + i

i + 1

=n∑

i=0

(−1)n−i

(n

i

)−

n∑i=0

(−1)n−i n

i + 1

(n

i

)

= 0−n∑

i=0

(−1)n−i n

n + 1

(n + 1i + 1

)= (−1)n+1 n

n + 1.

Third solution: (by Richard Stanley) Thecycle indica-tor of the symmetric groupSn is defined by

Zn(x1, . . . , xn) =∑

π∈Sn

xc1(π)1 · · ·xcn(π)

n ,

whereci(π) is the number of cycles ofπ of length i.Put

Fn =∑

π∈Sn

σ(π)xν(π) = Zn(x,−1, 1,−1, 1, . . . )

and

f(n) =∑

π∈Sn

σ(π)ν(π) + 1

=∫ 1

0

Fn(x) dx.


7

A standard argument in enumerative combinatorics (theExponential Formula) gives

∞∑n=0

Zn(x1, . . . , xn)tn

n!= exp

∞∑k=1

xktk

k,

yielding

∞∑n=0

f(n)tn

n!=∫ 1

0

exp(

xt− t2

2+

t3

3− · · ·

)dx

=∫ 1

0

e(x−1)t+log(1+t) dx

=∫ 1

0

(1 + t)e(x−1)t dx

=1t(1− e−t)(1 + t).

Expanding the right side as a Taylor series and compar-ing coefficients yields the desired result.

Fourth solution (sketch): (by David Savitt) We provethe identity of rational functions∑

π∈Sn

σ(π)ν(π) + x

=(−1)n+1n!(x + n− 1)x(x + 1) · · · (x + n)

by induction onn, which forx = 1 implies the desiredresult. (This can also be deduced as in the other solu-tions, but in this argument it is necessary to formulatethe strong induction hypothesis.)

Let R(n, x) be the right hand side of the above equa-tion. It is easy to verify that

R(x, n) = R(x + 1, n− 1) + (n− 1)!(−1)n+1

x

+n−1∑l=2

(−1)l−1 (n− 1)!(n− l)!

R(x, n− l),

since the sum telescopes. To prove the desired equality,it suffices to show that the left hand side satisfies thesame recurrence. This follows because we can classifyeachπ ∈ Sn as either fixingn, being ann-cycle, or hav-ing n in an l-cycle for one ofl = 2, . . . , n− 1; writingthe sum over these classes gives the desired recurrence.




A1 We change to cylindrical coordinates, i.e., we putr =√

x2 + y2. Then the given inequality is equivalent to

r2 + z2 + 8 ≤ 6r,

or

(r − 3)2 + z2 ≤ 1.

This defines a solid of revolution (a solid torus); thearea being rotated is the disc(x − 3)2 + z2 ≤ 1 inthexz-plane. By Pappus’s theorem, the volume of thisequals the area of this disc, which isπ, times the dis-tance through which the center of mass is being rotated,which is(2π)3. That is, the total volume is6π2.

A2 Suppose on the contrary that the setB of values ofn for which Bob has a winning strategy is finite; forconvenience, we includen = 0 in B, and writeB =b1, . . . , bm. Then for every nonnegative integern notin B, Alice must have some move on a heap ofn stonesleading to a position in which the second player wins.That is, every nonnegative integer not inB can be writ-ten asb+p−1 for someb ∈ B and some primep. How-ever, there are numerous ways to show that this cannothappen.

First solution: Let t be any integer bigger than all oftheb ∈ B. Then it is easy to write downt consecutivecomposite integers, e.g.,(t+1)!+2, . . . , (t+1)!+t+1.Taken = (t + 1)! + t; then for eachb ∈ B, n − b + 1is one of the composite integers we just wrote down.

Second solution: Let p1, . . . , p2m be any prime num-bers; then by the Chinese remainder theorem, there ex-ists a positive integerx such that

x − b1 ≡ −1 (mod p1pm+1)

. . .

x − bn ≡ −1 (mod pmp2m).

For eachb ∈ B, the unique integerp such thatx =b + p − 1 is divisible by at least two primes, and socannot itself be prime.

Third solution: (by Catalin Zara) Putb1 = 0, and taken = (b2 − 1) · · · (bm − 1); thenn is composite because3, 8 ∈ B, and for any nonzerob ∈ B, n − bi + 1 isdivisible by but not equal tobi − 1. (One could alsotaken = b2 · · · bm − 1, so thatn− bi +1 is divisible bybi.)

A3 We first observe that given any sequence of integersx1, x2, . . . satisfying a recursion

xk = f(xk−1, . . . , xk−n) (k > n),

wheren is fixed andf is a fixed polynomial ofn vari-ables with integer coefficients, for any positive integerN , the sequence moduloN is eventually periodic. Thisis simply because there are only finitely many possiblesequences ofn consecutive values moduloN , and oncesuch a sequence is repeated, every subsequent value isrepeated as well.

We next observe that if one can rewrite the same recur-sion as

xk−n = g(xk−n+1, . . . , xk) (k > n),

whereg is also a polynomial with integer coefficients,then the sequence extends uniquely to a doubly infinitesequence. . . , x−1, x0, x1, . . . which is fully periodicmodulo anyN . That is the case in the situation at hand,because we can rewrite the given recursion as

xk−2005 = xk+1 − xk.

It thus suffices to find 2005 consecutive terms divisibleby N in the doubly infinite sequence, for any fixedN(so in particular forN = 2006). Running the recursionbackwards, we easily find

x1 = x0 = · · · = x−2004 = 1

x−2005 = · · · = x−4009 = 0,

yielding the desired result.

A4 First solution: By the linearity of expectation, the av-erage number of local maxima is equal to the sum ofthe probability of having a local maximum atk overk = 1, . . . , n. Fork = 1, this probability is 1/2: giventhe pairπ(1), π(2), it is equally likely thatπ(1) orπ(2) is bigger. Similarly, fork = n, the probability is1/2. For1 < k < n, the probability is 1/3: given thepairπ(k− 1), π(k), π(k + 1), it is equally likely thatany of the three is the largest. Thus the average numberof local maxima is

2 · 1

2+ (n − 2) · 1

3=

n + 1

3.

Second solution: Another way to apply the linear-ity of expectation is to compute the probability thati ∈ 1, . . . , n occurs as a local maximum. The mostefficient way to do this is to imagine the permutation asconsisting of the symbols1, . . . , n, ∗ written in a circlein some order. The numberi occurs as a local maxi-mum if the two symbols it is adjacent to both belong tothe set∗, 1, . . . , i−1. There arei(i−1) pairs of suchsymbols andn(n−1) pairs in total, so the probability of


2

i occurring as a local maximum isi(i− 1)/(n(n− 1)),and the average number of local maxima is

n∑

i=1

i(i − 1)

n(n − 1)=

2

n(n − 1)

n∑

i=1

(

i

2

)

=2

n(n − 1)

(

n + 1

3

)

=n + 1

3.

One can obtain a similar (if slightly more intricate) so-lution inductively, by removing the known local maxi-mumn and splitting into two shorter sequences.

Remark: The usual term for a local maximum in thissense is apeak. The complete distribution for the num-ber of peaks is known; Richard Stanley suggests the ref-erence: F. N. David and D. E. Barton,CombinatorialChance, Hafner, New York, 1962, p. 162 and subse-quent.

A5 Since the desired expression involves symmetric func-tions of a1, . . . , an, we start by finding a polynomialwith a1, . . . , an as roots. Note that

1 ± i tan θ = e±iθ sec θ

so that

1 + i tan θ = e2iθ(1 − i tan θ).

Consequently, if we putω = e2inθ, then the polynomial

Qn(x) = (1 + ix)n − ω(1 − ix)n

has among its rootsa1, . . . , an. Since these are distinctandQn has degreen, these must be exactly the roots.

If we write

Qn(x) = cnxn + · · · + c1x + c0,

then a1 + · · · + an = −cn−1/cn and a1 · · · an =−c0/cn, so the ratio we are seeking iscn−1/c0. Byinspection,

cn−1 = nin−1 − ωn(−i)n−1 = nin−1(1 − ω)

c0 = 1 − ω

so

a1 + · · · + an

a1 · · ·an=

n n ≡ 1 (mod 4)

−n n ≡ 3 (mod 4).

Remark: The same argument shows that the ratio be-tween any twoodd elementary symmetric functions ofa1, . . . , an is independent ofθ.

A6 First solution: (by Daniel Kane) The probability is1 − 35

12π2 . We start with some notation and simplifi-cations. For simplicity, we assume without loss of gen-erality that the circle has radius 1. LetE denote the

expected value of a random variable over all choices ofP, Q, R. Write [XY Z] for the area of triangleXY Z.

If P, Q, R, S are the four points, we may ignore the casewhere three of them are collinear, as this occurs withprobability zero. Then the only way they can fail toform the vertices of a convex quadrilateral is if one ofthem lies inside the triangle formed by the other three.There are four such configurations, depending on whichpoint lies inside the triangle, and they are mutually ex-clusive. Hence the desired probability is 1 minus fourtimes the probability thatS lies inside trianglePQR.That latter probability is simplyE([PQR]) divided bythe area of the disc.

Let O denote the center of the circle, and letP ′, Q′, R′

be the projections ofP, Q, R onto the circle fromO.We can write

[PQR] = ±[OPQ] ± [OQR] ± [ORP ]

for a suitable choice of signs, determined as follows. Ifthe pointsP ′, Q′, R′ lie on no semicircle, then all of thesigns are positive. IfP ′, Q′, R′ lie on a semicircle inthat order andQ lies inside the triangleOPR, then thesign on[OPR] is positive and the others are negative.If P ′, Q′, R′ lie on a semicircle in that order andQ liesoutside the triangleOPR, then the sign on[OPR] isnegative and the others are positive.

We first calculate

E([OPQ] + [OQR] + [ORP ]) = 3E([OPQ]).

Write r1 = OP, r2 = OQ, θ = ∠POQ, so that

[OPQ] =1

2r1r2(sin θ).

The distribution ofr1 is given by2r1 on [0, 1] (e.g.,by the change of variable formula to polar coordinates),and similarly forr2. The distribution ofθ is uniform on[0, π]. These three distributions are independent; hence

E([OPQ])

=1

2

(∫ 1

0

2r2 dr

)2(1

π

∫ π

0

sin(θ) dθ

)

=4

9π,

and

E([OPQ] + [OQR] + [ORP ]) =4

3π.

We now treat the case whereP ′, Q′, R′ lie on a semicir-cle in that order. Putθ1 = ∠POQ andθ2 = ∠QOR;then the distribution ofθ1, θ2 is uniform on the region

0 ≤ θ1, 0 ≤ θ2, θ1 + θ2 ≤ π.

In particular, the distribution onθ = θ1 + θ2 is 2θπ2 on

[0, π]. PutrP = OP, rQ = OQ, rR = OR. Again, the


3

distribution onrP is given by2rP on [0, 1], and simi-larly for rQ, rR; these are independent from each otherand from the joint distribution ofθ1, θ2. Write E′(X)for the expectation of a random variableX restricted tothis part of the domain.

Let χ be the random variable with value 1 ifQ is insidetriangleOPR and 0 otherwise. We now compute

E′([OPR])

=1

2

(∫ 1

0

2r2 dr

)2(∫ π

0

2θ

π2sin(θ) dθ

)

=4

9πE′(χ[OPR])

= E′(2[OPR]2/θ)

=1

2

(∫ 1

0

2r3 dr

)2(∫ π

0

2θ

π2θ−1 sin2(θ) dθ

)

=1

8π.

Also recall that given any triangleXY Z, if T is chosenuniformly at random insideXY Z, the expectation of[TXY ] is the area of triangle bounded byXY and thecentroid ofXY Z, namely1

3 [XY Z].

Let χ be the random variable with value 1 ifQ is insidetriangleOPR and 0 otherwise. Then

E′([OPQ] + [OQR] + [ORP ] − [PQR])

= 2E′(χ([OPQ] + [OQR]) + 2E′((1 − χ)[OPR])

= 2E′(2

3χ[OPR]) + 2E′([OPR]) − 2E′(χ[OPR])

= 2E′([OPR]) − 2

3E′(χ[OPR]) =

29

36π.

Finally, note that the case whenP ′, Q′, R′ lie on a semi-circle in some order occurs with probability3/4. (Thecase where they lie on a semicircle proceeding clock-wise fromP ′ to its antipode has probability 1/4; thiscase and its two analogues are exclusive and exhaus-tive.) Hence

E([PQR])

= E([OPQ] + [OQR] + [ORP ])

− 3

4E′([OPQ] + [OQR] + [ORP ] − [PQR])

=4

3π− 29

48π=

35

48π,

so the original probability is

1 − 4E([PQR])

π= 1 − 35

12π2.

Second solution: (by David Savitt) As in the first so-lution, it suffices to check that forP, Q, R chosen uni-formly at random in the disc,E([PQR]) = 35

48π . Draw

the linesPQ, QR, RP , which with probability 1 di-vide the interior of the circle into seven regions. Puta = [PQR], let b1, b2, b3 denote the areas of the threeother regions sharing a side with the triangle, and letc1, c2, c3 denote the areas of the other three regions.Put A = E(a), B = E(b1), C = E(c1), so thatA + 3B + 3C = π.

Note thatc1 + c2 + c3 + a is the area of the regionin which we can choose a fourth pointS so that thequadrilateralPQRS fails to be convex. By comparingexpectations, we have3C + A = 4A, soA = C and4A + 3B = π.

We will computeB + 2A = B + 2C, which is the ex-pected area of the part of the circle cut off by a chordthrough two random pointsD, E, on the side of thechord not containing a third random pointF . Let h bethe distance from the centerO of the circle to the lineDE. We now determine the distribution ofh.

Putr = OD; the distribution ofr is 2r on [0, 1]. With-out loss of generality, supposeO is the origin andDlies on the positivex-axis. For fixedr, the distributionof h runs over[0, r], and can be computed as the areaof the infinitesimal region in whichE can be chosen sothe chord throughDE has distance toO betweenh andh + dh, divided byπ. This region splits into two sym-metric pieces, one of which lies between chords makingangles ofarcsin(h/r) andarcsin((h + dh)/r) with thex-axis. The angle between these isdθ = dh/(r2 − h2).Draw the chord throughD at distanceh to O, and letL1, L2 be the lengths of the parts on opposite sides ofD; then the area we are looking for is12 (L2

1 + L22)dθ.

Since

L1, L2 =√

1 − h2 ±√

r2 − h2,

the area we are seeking (after doubling) is

21 + r2 − 2h2

√r2 − h2

.

Dividing by π, then integrating overr, we compute thedistribution ofh to be

1

π

∫ 1

h

21 + r2 − 2h2

√r2 − h2

2r dr

=16

3π(1 − h2)3/2.

We now return to computingB + 2A. Let A(h) de-note the smaller of the two areas of the disc cut off bya chord at distanceh. The chance that the third pointis in the smaller (resp. larger) portion isA(h)/π (resp.1−A(h)/π), and then the area we are trying to computeis π − A(h) (resp.A(h)). Using the distribution onh,and the fact that

A(h) = 2

∫ 1

h

√

1 − h2 dh

=π

2− arcsin(h) − h

√

1 − h2,


4

we find

B + 2A

=2

π

∫ 1

0

A(h)(π − A(h))16

3π(1 − h2)3/2 dh

=35 + 24π2

72π.

Since4A + 3B = π, we solve to obtainA = 3548π as in

the first solution.

Third solution: (by Noam Elkies) Again, we reduceto computing the average area of a triangle formedby three random pointsA, B, C inside a unit cir-cle. Let O be the center of the circle, and putc =maxOA, OB, OC; then the probability thatc ≤ ris (r2)3, so the distribution ofc is 6c5 dc on [0, 1].

Givenc, the expectation of[ABC] is equal toc2 timesX , the expected area of a triangle formed by two ran-dom pointsP, Q in a circle and a fixed pointR onthe boundary. We introduce polar coordinates centeredat R, in which the circle is given byr = 2 sin θ forθ ∈ [0, π]. The distribution of a random point in thatcircle is 1

π r dr dθ over θ ∈ [0, π] andr ∈ [0, 2 sin θ].If (r, θ) and(r′, θ′) are the two random points, then thearea is1

2rr′ sin |θ − θ′|.Performing the integrals overr andr′ first, we find

X =32

9π2

∫ π

0

∫ π

0

sin3 θ sin3 θ′ sin |θ − θ′| dθ′ dθ

=64

9π2

∫ π

0

∫ θ

0

sin3 θ sin3 θ′ sin(θ − θ′) dθ′ dθ.

This integral is unpleasant but straightforward; it yieldsX = 35/(36π), and E([PQR]) =

∫ 1

0 6c7X dc =35/(48π), giving the desired result.

Remark: This is one of the oldest problems in geo-metric probability; it is an instance of Sylvester’s four-point problem, which nowadays is usually solved us-ing a device known as Crofton’s formula. We defer tohttp://mathworld.wolfram.com/ for furtherdiscussion.

B1 The “curve”x3+3xy+y3−1 = 0 is actually reducible,because the left side factors as

(x + y − 1)(x2 − xy + y2 + x + y + 1).

Moreover, the second factor is

1

2((x + 1)2 + (y + 1)2 + (x − y)2),

so it only vanishes at(−1,−1). Thus the curve in ques-tion consists of the single point(−1,−1) together withthe linex + y = 1. To form a triangle with three pointson this curve, one of its vertices must be(−1,−1). Theother two vertices lie on the linex + y = 1, so the

length of the altitude from(−1,−1) is the distance from(−1,−1) to (1/2, 1/2), or3

√2/2. The area of an equi-

lateral triangle of heighth is h2√

3/3, so the desiredarea is3

√3/2.

Remark: The factorization used above is a special caseof the fact that

x3 + y3 + z3 − 3xyz

= (x + y + z)(x + ωy + ω2z)(x + ω2y + ωz),

whereω denotes a primitive cube root of unity. Thatfact in turn follows from the evaluation of the determi-nant of thecirculant matrix

x y zz x yy z x

by reading off the eigenvalues of the eigenvectors(1, ωi, ω2i) for i = 0, 1, 2.

B2 Letx = x − ⌊x⌋ denote the fractional part ofx. Fori = 0, . . . , n, putsi = x1 + · · · + xi (so thats0 = 0).Sort the numberss0, . . . , sn into ascending order,and call the resultt0, . . . , tn. Since0 = t0 ≤ · · · ≤tn < 1, the differences

t1 − t0, . . . , tn − tn−1, 1 − tn

are nonnegative and add up to 1. Hence (as in the pi-geonhole principle) one of these differences is no morethan1/(n + 1); if it is anything other than1 − tn, itequals±(si − sj) for some0 ≤ i < j ≤ n. PutS = xi+1, . . . , xj andm = ⌊si⌋ − ⌊sj⌋; then

∣

∣

∣

∣

∣

m +∑

s∈S

s

∣

∣

∣

∣

∣

= |m + sj − si|

= |sj − si|

≤ 1

n + 1,

as desired. In case1 − tn ≤ 1/(n + 1), we takeS =x1, . . . , xn and m = −⌈sn⌉, and again obtain thedesired conclusion.

B3 The maximum is(

n2

)

+ 1, achieved for instance by aconvexn-gon: besides the trivial partition (in which allof the points are in one part), each linear partition oc-curs by drawing a line crossing a unique pair of edges.

First solution: We will prove thatLS =(

n2

)

+1 in anyconfiguration in which no two of the lines joining pointsof S are parallel. This suffices to imply the maximumin all configurations: given a maximal configuration,we may vary the points slightly to get another maximalconfiguration in which our hypothesis is satisfied. Forconvenience, we assumen ≥ 3, as the casesn = 1, 2are easy.


5

Let P be the line at infinity in the real projective plane;i.e., P is the set of possible directions of lines in theplane, viewed as a circle. Remove the directions corre-sponding to lines through two points ofS; this leavesbehind

(

n2

)

intervals.

Given a direction in one of the intervals, consider theset of linear partitions achieved by lines parallel to thatdirection. Note that the resulting collection of partitionsdepends only on the interval. Then note that the collec-tions associated to adjacent intervals differ in only oneelement.

The trivial partition that puts all ofS on one side isin every such collection. We now observe that for anyother linear partitionA, B, the set of intervals towhichA, B is:

(a) a consecutive block of intervals, but

(b) not all of them.

For (a), note that ifℓ1, ℓ2 are nonparallel lines achievingthe same partition, then we can rotate around their pointof intersection to achieve all of the intermediate direc-tions on one side or the other. For (b), the casen = 3is evident; to reduce the general case to this case, takepointsP, Q, R such thatP lies on the opposite side ofthe partition fromQ andR.

It follows now that that each linear partition, except forthe trivial one, occurs in exactly one place as the parti-tion associated to some interval but not to its immediatecounterclockwise neighbor. In other words, the num-ber of linear partitions is one more than the number ofintervals, or

(

n2

)

+ 1 as desired.

Second solution: We prove the upper bound by induc-tion on n. Choose a pointP in the convex hull ofS.PutS′ = S \ P; by the induction hypothesis, thereare at most

(

n−12

)

+ 1 linear partitions ofS′. Note thateach linear partition ofS restricts to a linear partitionof S′. Moreover, if two linear partitions ofS restrict tothe same linear partition ofS′, then that partition ofS′

is achieved by a line throughP .

By rotating a line throughP , we see that there are atmostn − 1 partitions ofS′ achieved by lines throughP : namely, the partition only changes when the rotatingline passes through one of the points ofS. This yieldsthe desired result.

Third solution: (by Noam Elkies) We enlarge the planeto a projective plane by adding a line at infinity, thenapply the polar duality map centered at one of the pointsO ∈ S. This turns the rest ofS into a setS′ of n − 1lines in the dual projective plane. LetO′ be the pointin the dual plane corresponding to the original line atinfinity; it does not lie on any of the lines inS′.

Let ℓ be a line in the original plane, corresponding to apointP in the dual plane. If we form the linear partitioninduced byℓ, then the points ofS \ O lying in thesame part asO correspond to the lines ofS′ which cross

the segmentO′P . If we consider the dual affine planeas being divided into regions by the lines ofS′, then thelines ofS′ crossing the segmentO′P are determined bywhich regionP lies in.

Thus our original maximum is equal to the maximumnumber of regions into whichn−1 lines divide an affineplane. By induction onn, this number is easily seen tobe1 +

(

n2

)

.

Remark: Given a finite setS of points inRn, a non-Radon partition of S is a pair(A, B) of complementarysubsets that can be separated by a hyperplane.Radon’stheorem states that if#S ≥ n+2, then not every(A, B)is a non-Radon partition. The result of this problem hasbeen greatly extended, especially within the context ofmatroid theory and oriented matroid theory. RichardStanley suggests the following references: T. H. Bry-lawski, A combinatorial perspective on the Radon con-vexity theorem,Geom. Ded. 5 (1976), 459-466; and T.Zaslavsky, Extremal arrangements of hyperplanes,Ann.N. Y. Acad. Sci. 440 (1985), 69-87.

B4 The maximum is2k, achieved for instance by the sub-space

(x1, . . . , xn) ∈ Rn : x1 = · · · = xn−k = 0.

First solution: More generally, we show that any affinek-dimensional plane inRn can contain at most2k

points inZ. The proof is by induction onk + n; thecasek = n = 0 is clearly true.

Suppose thatV is ak-plane inRn. Denote the hyper-planesxn = 0 andxn = 1 by V0 andV1, respec-tively. If V ∩ V0 andV ∩ V1 are each at most(k − 1)-dimensional, thenV ∩V0∩Z andV ∩V1∩Z each havecardinality at most2k−1 by the induction assumption,and henceV ∩ Z has at most2k elements. Otherwise,if V ∩ V0 or V ∩ V1 is k-dimensional, thenV ⊂ V0

or V ⊂ V1; now apply the induction hypothesis onV ,viewed as a subset ofRn−1 by dropping the last coor-dinate.

Second solution: Let S be a subset ofZ contained ina k-dimensional subspace ofV . This is equivalent toasking that anyt1, . . . , tk+1 ∈ S satisfy a nontriviallinear dependencec1t1 + · · · + ck+1tk+1 = 0 withc1, . . . , ck+1 ∈ R. Sincet1, . . . , tk+1 ∈ Qn, givensuch a dependence we can always find another one withc1, . . . , ck+1 ∈ Q; then by clearing denominators, wecan find one withc1, . . . , ck+1 ∈ Z and not all having acommon factor.

Let F2 denote the field of two elements, and letS ⊆ Fn2

be the reductions modulo 2 of the points ofS. Then anyt1, . . . , tk+1 ∈ S satisfy a nontrivial linear dependence,because we can take the dependence from the end ofthe previous paragraph and reduce modulo 2. HenceSis contained in ak-dimensional subspace ofF2n , andthe latter has cardinality exactly2k. ThusS has at most2k elements, as doesS.


6

Variant (suggested by David Savitt): ifS containedk +1 linearly independent elements, the(k +1)×n matrixformed by these would have a nonvanishing maximalminor. The lift of that minor back toR would also notvanish, soS would containk + 1 linearly independentelements.

Third solution: (by Catalin Zara) LetV be a k-dimensional subspace. Form the matrix whose rows arethe elements ofV ∩Z; by construction, it has row rankat mostk. It thus also has column rank at mostk; inparticular, we can choosek coordinates such that eachpoint of V ∩ Z is determined by thosek of its coordi-nates. Since each coordinate of a point inZ can onlytake two values,V ∩ Z can have at most2k elements.

Remark: The proposers probably did notrealize that this problem appeared onlineabout three months before the exam, athttp://www.artofproblemsolving.com/Forum/viewtopic.php?t=105991. (It mayvery well have also appeared even earlier.)

B5 The answer is1/16. We have

∫ 1

0

x2f(x) dx −∫ 1

0

xf(x)2 dx

=

∫ 1

0

(x3/4 − x(f(x) − x/2)2) dx

≤∫ 1

0

x3/4 dx = 1/16,

with equality whenf(x) = x/2.

B6 First solution: We start with some easy upper andlower bounds onan. We writeO(f(n)) andΩ(f(n))for functionsg(n) such thatf(n)/g(n) andg(n)/f(n),respectively, are bounded above. Sincean is a non-decreasing sequence,an+1 − an is bounded above, soan = O(n). That meansa−1/k

n = Ω(n−1/k), so

an = Ω

(

n∑

i=1

i−1/k

)

= Ω(n(k−1)/k).

In fact, all we will need is thatan → ∞ asn → ∞.

By Taylor’s theorem with remainder, for1 < m < 2andx > 0,

|(1 + x)m − 1 − mx| ≤ m(m − 1)

2x2.

Taking m = (k + 1)/k and x = an+1/an = 1 +

a−(k+1)/kn , we obtain

∣

∣

∣

∣

a(k+1)/kn+1 − a(k+1)/k

n − k + 1

k

∣

∣

∣

∣

≤ k + 1

2k2a−(k+1)/k

n .

In particular,

limn→∞

a(k+1)/kn+1 − a(k+1)/k

n =k + 1

k.

In general, ifxn is a sequence withlimn→∞ xn = c,then also

limn→∞

1

n

n∑

i=1

xi = c

by Cesaro’s lemma. Explicitly, for anyǫ > 0, we canfind N such that|xn − c| ≤ ǫ/2 for n ≥ N , and then

∣

∣

∣

∣

∣

c − 1

n

n∑

i=1

xi

∣

∣

∣

∣

∣

≤ n − N

n

ǫ

2+

N

n

∣

∣

∣

∣

∣

N∑

i=1

(c − xi)

∣

∣

∣

∣

∣

;

for n large, the right side is smaller thanǫ.

In our case, we deduce that

limn→∞

a(k+1)/kn

n=

k + 1

k

and so

limn→∞

ak+1n

nk=

(

k + 1

k

)k

,

as desired.

Remark: The use of Cesaro’s lemma above is the spe-cial casebn = n of theCesaro-Stolz theorem: if an, bn

are sequences such thatbn is positive, strictly increas-ing, and unbounded, and

limn→∞

an+1 − an

bn+1 − bn= L,

then

limn→∞

an

bn= L.

Second solution: In this solution, rather than applyingTaylor’s theorem with remainder to(1 + x)m for 1 <m < 2 andx > 0, we only apply convexity to deducethat(1 + x)m ≥ 1 + mx. This gives

a(k+1)/kn+1 − a(k+1)/k

n ≥ k + 1

k,

and so

a(k+1)/kn ≥ k + 1

kn + c

for somec ∈ R. In particular,

lim infn→∞

a(k+1)/kn

n≥ k + 1

k

and so

lim infn→∞

an

nk/(k+1)≥(

k + 1

k

)k/(k+1)

.


7

But turning this around, the fact that

an+1 − an

= a−1/kn

≤(

k + 1

k

)−1/(k+1)

n−1/(k+1)(1 + o(1)),

whereo(1) denotes a function tending to 0 asn → ∞,yields

an

≤(

k + 1

k

)−1/(k+1) n∑

i=1

i−1/(k+1)(1 + o(1))

=k + 1

k

(

k + 1

k

)−1/(k+1)

nk/(k+1)(1 + o(1))

=

(

k + 1

k

)k/(k+1)

nk/(k+1)(1 + o(1)),

so

lim supn→∞

an

nk/(k+1)≤(

k + 1

k

)k/(k+1)

and this completes the proof.

Third solution: We argue thatan → ∞ as in the firstsolution. Writebn = an − Lnk/(k+1), for a value ofLto be determined later. We have

bn+1

= bn + a−1/kn − L((n + 1)k/(k+1) − nk/(k+1))

= e1 + e2,

where

e1 = bn + a−1/kn − L−1/kn−1/(k+1)

e2 = L((n + 1)k/(k+1) − nk/(k+1))

− L−1/kn−1/(k+1).

We first estimatee1. For−1 < m < 0, by the convexityof (1 + x)m and(1 + x)1−m, we have

1 + mx ≤ (1 + x)m

≤ 1 + mx(1 + x)m−1.

Hence

−1

kL−(k+1)/kn−1bn ≤ e1 − bn

≤ −1

kbna−(k+1)/k

n .

Note that both bounds have sign opposite tobn; more-over, by the boundan = Ω(n(k−1)/k), both boundshave absolutely value strictly less than that ofbn for nsufficiently large. Consequently, forn large,

|e1| ≤ |bn|.We now work one2. By Taylor’s theorem with remain-der applied to(1 + x)m for x > 0 and0 < m < 1,

1 + mx ≥ (1 + x)m

≥ 1 + mx +m(m − 1)

2x2.

The “main term” of L((n + 1)k/(k+1) − nk/(k+1))is L k

k+1n−1/(k+1). To make this coincide with

L−1/kn−1/(k+1), we take

L =

(

k + 1

k

)k/(k+1)

.

We then find that

|e2| = O(n−2),

and becausebn+1 = e1 + e2, we have|bn+1| ≤ |bn| +|e2|. Hence

|bn| = O

(

n∑

i=1

i−2

)

= O(1),

and so

limn→∞

ak+1n

nk= Lk+1 =

(

k + 1

k

)k

.

Remark: The casek = 2 appeared on the 2004 Roma-nian Olympiad (district level).

Remark: One can make a similar argument for any se-quence given byan+1 = an + f(an), whenf is ade-creasing function.

Remark: Richard Stanley suggests a heuristic for de-termining the asymptotic behavior of sequences of thistype: replace the given recursion

an+1 − an = a−1/kn

by the differential equation

y′ = y−1/k

and determine the asymptotics of the latter.




A1 The only suchα are2/3, 3/2, (13±√

601)/12.

First solution: Let C1 andC2 be the curvesy = αx2 +αx + 1

24 andx = αy2 + αy + 124 , respectively, and let

L be the liney = x. We consider three cases.

If C1 is tangent toL, then the point of tangency(x, x)satisfies

2αx + α = 1, x = αx2 + αx +1

24;

by symmetry,C2 is tangent toL there, soC1 andC2 aretangent. Writingα = 1/(2x + 1) in the first equationand substituting into the second, we must have

x =x2 + x

2x + 1+

1

24,

which simplifies to0 = 24x2−2x−1 = (6x+1)(4x−1), orx ∈ 1/4,−1/6. This yieldsα = 1/(2x + 1) ∈2/3, 3/2.

If C1 does not intersectL, thenC1 andC2 are separatedby L and so cannot be tangent.

If C1 intersectsL in two distinct pointsP1, P2, then itis not tangent toL at either point. Suppose at one ofthese points, sayP1, the tangent toC1 is perpendicularto L; then by symmetry, the same will be true ofC2, soC1 andC2 will be tangent atP1. In this case, the pointP1 = (x, x) satisfies

2αx + α = −1, x = αx2 + αx +1

24;

writing α = −1/(2x + 1) in the first equation and sub-stituting into the second, we have

x = −x2 + x

2x + 1+

1

24,

or x = (−23 ±√

601)/72. This yieldsα = −1/(2x +

1) = (13 ±√

601)/12.

If instead the tangents toC1 at P1, P2 are not perpen-dicular toL, then we claim there cannot be any pointwhereC1 andC2 are tangent. Indeed, if we count inter-sections ofC1 andC2 (by usingC1 to substitute foryin C2, then solving fory), we get at most four solutionscounting multiplicity. Two of these areP1 andP2, andany point of tangency counts for two more. However,off of L, any point of tangency would have a mirror im-age which is also a point of tangency, and there cannotbe six solutions. Hence we have now found all possibleα.

Second solution: For any nonzero value ofα, the twoconics will intersect in four points in the complex pro-jective planeP2(C). To determine they-coordinates ofthese intersection points, subtract the two equations toobtain

(y − x) = α(x − y)(x + y) + α(x − y).

Therefore, at a point of intersection we have eitherx = y, or x = −1/α − (y + 1). Substituting thesetwo possible linear conditions into the second equationshows that they-coordinate of a point of intersection isa root of eitherQ1(y) = αy2 + (α − 1)y + 1/24 orQ2(y) = αy2 + (α + 1)y + 25/24 + 1/α.

If two curves are tangent, then they-coordinates of atleast two of the intersection points will coincide; theconverse is also true because one of the curves is thegraph of a function inx. The coincidence occurs pre-cisely when either the discriminant of at least one ofQ1 or Q2 is zero, or there is a common root ofQ1 andQ2. Computing the discriminants ofQ1 andQ2 yields(up to constant factors)f1(α) = 6α2 − 13α + 6 andf2(α) = 6α2 − 13α − 18, respectively. If on the otherhandQ1 andQ2 have a common root, it must be alsoa root ofQ2(y) − Q1(y) = 2y + 1 + 1/α, yieldingy = −(1 + α)/(2α) and0 = Q1(y) = −f2(α)/(24α).

Thus the values ofα for which the two curves are tan-gent must be contained in the set of zeros off1 andf2,namely2/3, 3/2, and(13 ±

√601)/12.

Remark: The fact that the two conics inP2(C) meet infour points, counted with multiplicities, is a special caseof Bezout’s theorem: two curves inP2(C) of degreesm, n and not sharing any common component meet inexactlymn points when counted with multiplicity.

Many solvers were surprised that the proposers chosethe parameter1/24 to give two rational roots and twononrational roots. In fact, they had no choice in thematter: attempting to make all four roots rational byreplacing1/24 by β amounts to asking forβ2 + β andβ2 + β + 1 to be perfect squares. This cannot happenoutside of trivial cases (β = 0,−1) ultimately becausethe elliptic curve 24A1 (in Cremona’s notation) overQ

has rank0. (Thanks to Noam Elkies for providing thiscomputation.)

However, there are choices that make the radical milder,e.g.,β = 1/3 givesβ2 + β = 4/9 andβ2 + β + 1 =13/9, while β = 3/5 givesβ2 + β = 24/25 andβ2 +β + 1 = 49/25.

A2 The minimum is 4, achieved by the square with vertices(±1,±1).


2

First solution: To prove that 4 is a lower bound,let S be a convex set of the desired form. ChooseA, B, C, D ∈ S lying on the branches of the two hyper-bolas, withA in the upper right quadrant,B in the upperleft, C in the lower left,D in the lower right. Then thearea of the quadrilateralABCD is a lower bound forthe area ofS.

Write A = (a, 1/a), B = (b,−1/b), C = (−c,−1/c),D = (−d, 1/d) with a, b, c, d > 0. Then the area of thequadrilateralABCD is

1

2(a/b + b/c + c/d + d/a + b/a + c/b + d/c + a/d),

which by the arithmetic-geometric mean inequality is atleast4.

Second solution: ChooseA, B, C, D as in the first so-lution. Note that both the hyperbolas and the area ofthe convex hull ofABCD are invariant under the trans-formation(x, y) 7→ (xm, y/m) for any m > 0. Form small, the counterclockwise angle from the lineACto the lineBD approaches 0; form large, this angleapproachesπ. By continuity, for somem this angle be-comesπ/2, that is,AC andBD become perpendicular.The area ofABCD is thenAC · BD.

It thus suffices to note thatAC ≥ 2√

2 (and similarlyfor BD). This holds because if we draw the tangentlines to the hyperbolaxy = 1 at the points(1, 1) and(−1,−1), thenA andC lie outside the region betweenthese lines. If we project the segmentAC orthogonallyonto the linex = y = 1, the resulting projection haslength at least2

√2, soAC must as well.

Third solution: (by Richard Stanley) ChooseA, B, C, D as in the first solution. Now fixingA andC, moveB andD to the points at which the tangentsto the curve are parallel to the lineAC. This does notincrease the area of the quadrilateralABCD (even ifthis quadrilateral is not convex).

Note thatB and D are now diametrically opposite;write B = (−x, 1/x) andD = (x,−1/x). If we thusrepeat the procedure, fixingB andD and movingA andC to the points where the tangents are parallel toBD,thenA andC must move to(x, 1/x) and(−x,−1/x),respectively, forming a rectangle of area 4.

Remark: Many geometric solutions are possible. Anexample suggested by David Savitt (due to ChrisBrewer): note thatAD andBC cross the positive andnegativex-axes, respectively, so the convex hull ofABCD containsO. Then check that the area of tri-angleOAB is at least 1, et cetera.

A3 Assume that we have an ordering of1, 2, . . . , 3k +1 such that no initial subsequence sums to0 mod3. If we omit the multiples of3 from this order-ing, then the remaining sequence mod3 must looklike 1, 1,−1, 1,−1, . . . or −1,−1, 1,−1, 1, . . .. Sincethere is one more integer in the ordering congruent to1

mod3 than to−1, the sequence mod3 must look like1, 1,−1, 1,−1, . . ..

It follows that the ordering satisfies the given conditionif and only if the following two conditions hold: thefirst element in the ordering is not divisible by3, andthe sequence mod3 (ignoring zeroes) is of the form1, 1,−1, 1,−1, . . .. The two conditions are indepen-dent, and the probability of the first is(2k+1)/(3k+1)

while the probability of the second is1/(

2k+1k

)

, sincethere are

(

2k+1k

)

ways to order(k + 1) 1’s andk −1’s.Hence the desired probability is the product of thesetwo, or k!(k+1)!

(3k+1)(2k)! .

A4 Note thatn is a repunit if and only if9n + 1 = 10m forsome power of 10 greater than 1. Consequently, if weput

g(n) = 9f

(

n − 1

9

)

+ 1,

thenf takes repunits to repunits if and only ifg takespowers of 10 greater than 1 to powers of 10 greater than1. We will show that the only such functionsg are thoseof the formg(n) = 10cnd for d ≥ 0, c ≥ 1 − d (all ofwhich clearly work), which will mean that the desiredpolynomialsf are those of the form

f(n) =1

9(10c(9n + 1)d − 1)

for the samec, d.

It is convenient to allow “powers of 10” to be of theform 10k for any integerk. With this convention, itsuffices to check that the polynomialsg taking powersof 10 greater than 1 to powers of 10 are of the form10cnd for any integersc, d with d ≥ 0.

First solution: Suppose that the leading term ofg(x)is axd, and note thata > 0. As x → ∞, we haveg(x)/xd → a; however, forx a power of 10 greaterthan 1,g(x)/xd is a power of 10. The set of powers of10 has no positive limit point, sog(x)/xd must be equalto a for x = 10k with k sufficiently large, and we musthavea = 10c for somec. The polynomialg(x)−10cxd

has infinitely many roots, so must be identically zero.

Second solution: We proceed by induction ond =deg(g). If d = 0, we haveg(n) = 10c for somec.Otherwise,g has rational coefficients by Lagrange’s in-terpolation formula (this applies to any polynomial ofdegreed taking at leastd+1 different rational numbersto rational numbers), sog(0) = t is rational. More-over,g takes each value only finitely many times, so thesequenceg(100), g(101), . . . includes arbitrarily largepowers of 10. Suppose thatt 6= 0; then we can choosea positive integerh such that the numerator oft is notdivisible by10h. But for c large enough,g(10c)− t hasnumerator divisible by10b for someb > h, contradic-tion.


3

Consequently,t = 0, and we may apply the inductionhypothesis tog(n)/n to deduce the claim.

Remark: The second solution amounts to the fact thatg, being a polynomial with rational coefficients, is con-tinuous for the2-adic and5-adic topologies onQ. Bycontrast, the first solution uses the “∞-adic” topology,i.e., the usual real topology.

A5 In all solutions, letG be a finite group of orderm.

First solution: By Lagrange’s theorem, ifm is notdivisible by p, then n = 0. Otherwise, letS bethe set ofp-tuples (a0, . . . , ap−1) ∈ Gp such thata0 · · ·ap−1 = e; thenS has cardinalitymp−1, whichis divisible byp. Note that this set is invariant undercyclic permutation, that is, if(a0, . . . , ap−1) ∈ S, then(a1, . . . , ap−1, a0) ∈ S also. The fixed points underthis operation are the tuples(a, . . . , a) with ap = e; allother tuples can be grouped into orbits under cyclic per-mutation, each of which has sizep. Consequently, thenumber ofa ∈ G with ap = e is divisible byp; sincethat number isn + 1 (only e has order 1), this provesthe claim.

Second solution: (by Anand Deopurkar) Assume thatn > 0, and letH be any subgroup ofG of orderp. LetS be the set of all elements ofG\H of order dividingp,and letH act onG by conjugation. Each orbit has sizep except for those which consist of individual elementsg which commute withH . For each suchg, g andHgenerate an elementary abelian subgroup ofG of orderp2. However, we can group theseg into sets of sizep2 − p based on which subgroup they generate togetherwith H . Hence the cardinality ofS is divisible byp;adding thep−1 nontrivial elements ofH givesn ≡ −1(mod p) as desired.

Third solution: Let S be the set of elements inG hav-ing order dividingp, and letH be an elementary abelianp-group of maximal order inG. If |H | = 1, then weare done. So assume|H | = pk for somek ≥ 1, andlet H act onS by conjugation. LetT ⊂ S denotethe set of fixed points of this action. Then the sizeof every H-orbit on S dividespk, and so|S| ≡ |T |(mod p). On the other hand,H ⊂ T , and if T con-tained an element not inH , then that would contradictthe maximality ofH . It follows thatH = T , and so|S| ≡ |T | = |H | = pk ≡ 0 (mod p), i.e., |S| = n + 1is a multiple ofp.

Remark: This result is a theorem of Cauchy; the firstsolution above is due to McKay. A more general (andmore difficult) result was proved by Frobenius: for anypositive integerm, if G is a finite group of order divis-ible by m, then the number of elements ofG of orderdividing m is a multiple ofm.

A6 For an admissible triangulationT , number the verticesof P consecutivelyv1, . . . , vn, and letai be the numberof edges inT emanating fromvi; note thatai ≥ 2 forall i.

We first claim thata1 + · · ·+an ≤ 4n−6. LetV, E, Fdenote the number of vertices, edges, and faces inT .By Euler’s Formula,(F +1)−E+V = 2 (one must add1 to the face count for the region exterior toP ). Eachface has three edges, and each edge but then outsideedges belongs to two faces; henceF = 2E −n. On theother hand, each edge has two endpoints, and each oftheV − n internal vertices is an endpoint of at least6edges; hencea1+· · ·+an+6(V −n) ≤ 2E. Combiningthis inequality with the previous two equations gives

a1 + · · · + an ≤ 2E + 6n− 6(1 − F + E)

= 4n− 6,

as claimed.

Now setA3 = 1 andAn = An−1 + 2n − 3 for n ≥ 4;we will prove by induction onn thatT has at mostAn

triangles. Forn = 3, sincea1 + a2 + a3 = 6, a1 =a2 = a3 = 2 and henceT consists of just one triangle.

Next assume that an admissible triangulation of an(n − 1)-gon has at mostAn−1 triangles, and letTbe an admissible triangulation of ann-gon. If anyai = 2, then we can remove the triangle ofT contain-ing vertexvi to obtain an admissible triangulation of an(n−1)-gon; then the number of triangles inT is at mostAn−1 + 1 < An by induction. Otherwise, allai ≥ 3.Now the average ofa1, . . . , an is less than4, and thusthere are moreai = 3 than ai ≥ 5. It follows thatthere is a sequence ofk consecutive vertices inP whosedegrees are3, 4, 4, . . . , 4, 3 in order, for somek with2 ≤ k ≤ n− 1 (possiblyk = 2, in which case there areno degree4 vertices separating the degree3 vertices). Ifwe remove fromT the2k−1 triangles which contain atleast one of these vertices, then we are left with an ad-missible triangulation of an(n− 1)-gon. It follows thatthere are at mostAn−1+2k−1 ≤ An−1+2n−3 = An

triangles inT . This completes the induction step andthe proof.

Remark: We can refine the boundAn somewhat. Sup-posing thatai ≥ 3 for all i, the fact thata1 + · · ·+an ≤4n − 6 implies that there are at least six more indicesi with ai = 3 than withai ≥ 5. Thus there exist sixsequences with degrees3, 4, . . . , 4, 3, of total length atmostn + 6. We may thus choose a sequence of lengthk ≤ ⌊n

6 ⌋ + 1, so we may improve the upper bound toAn = An−1 + 2⌊n

6 ⌋ + 1, or asymptotically16n2.

However (as noted by Noam Elkies), a hexagonalswatch of a triangular lattice, with the boundary as closeto regular as possible, achieves asymptotically1

6n2 tri-angles.

B1 The problem fails iff is allowed to be constant, e.g.,takef(n) = 1. We thus assume thatf is nonconstant.


4

Write f(n) =∑d

i=0 aini with ai > 0. Then

f(f(n) + 1) =d

∑

i=0

ai(f(n) + 1)i

≡ f(1) (mod f(n)).

If n = 1, then this implies thatf(f(n) + 1) is divisibleby f(n). Otherwise,0 < f(1) < f(n) sincef is non-constant and has positive coefficients, sof(f(n) + 1)cannot be divisible byf(n).

B2 PutB = max0≤x≤1 |f ′(x)| andg(x) =∫ x

0f(y) dy.

Sinceg(0) = g(1) = 0, the maximum value of|g(x)|must occur at a critical pointy ∈ (0, 1) satisfyingg′(y) = f(y) = 0. We may thus takeα = y hereafter.

Since∫ α

0f(x) dx = −

∫ 1−α

0f(1 − x) dx, we may as-

sume thatα ≤ 1/2. By then substituting−f(x) forf(x) if needed, we may assume that

∫ α

0f(x) dx ≥ 0.

From the inequalityf ′(x) ≥ −B, we deducef(x) ≤B(α − x) for 0 ≤ x ≤ α, so

∫ α

0

f(x) dx ≤∫ α

0

B(α − x) dx

= − 1

2B(α − x)2

∣

∣

∣

∣

α

0

=α2

2B ≤ 1

8B

as desired.

B3 First solution: Observing thatx2/2 = 13, x3/4 = 34,x4/8 = 89, we guess thatxn = 2n−1F2n+3, whereFk is thek-th Fibonacci number. Thus we claim thatxn = 2n−1

√5

(α2n+3 − α−(2n+3)), whereα = 1+√

52 , to

make the answerx2007 = 22006

√5

(α3997 − α−3997).

We prove the claim by induction; the base casex0 =1 is true, and so it suffices to show that the recursionxn+1 = 3xn + ⌊xn

√5⌋ is satisfied for our formula for

xn. Indeed, sinceα2 = 3+√

52 , we have

xn+1 − (3 +√

5)xn =2n−1

√5

(2(α2n+5 − α−(2n+5))

− (3 +√

5)(α2n+3 − α−(2n+3)))

= 2nα−(2n+3).

Now 2nα−(2n+3) = (1−√

52 )3(3−

√5)n is between−1

and0; the recursion follows sincexn, xn+1 are integers.

Second solution: (by Catalin Zara) Sincexn is rational,we have0 < xn

√5 − ⌊xn

√5⌋ < 1. We now have the

inequalities

xn+1 − 3xn < xn

√5 < xn+1 − 3xn + 1

(3 +√

5)xn − 1 < xn+1 < (3 +√

5)xn

4xn − (3 −√

5) < (3 −√

5)xn+1 < 4xn

3xn+1 − 4xn < xn+1

√5 < 3xn+1 − 4xn + (3 −

√5).

Since0 < 3 −√

5 < 1, this yields⌊xn+1

√5⌋ =

3xn+1−4xn, so we can rewrite the recursion asxn+1 =6xn −4xn−1 for n ≥ 2. It is routine to solve this recur-sion to obtain the same solution as above.

Remark: With an initial 1 prepended, thisbecomes sequence A018903 in Sloane’s On-Line Encyclopedia of Integer Sequences:(http://www.research.att.com/˜njas/sequences/ ). Therein, the sequence is describedas the caseS(1, 5) of the sequenceS(a0, a1)in which an+2 is the least integer for whichan+2/an+1 > an+1/an. Sloane cites D. W. Boyd,Linear recurrence relations for some generalized Pisotsequences,Advances in Number Theory (Kingston,ON, 1991), Oxford Univ. Press, New York, 1993, p.333–340.

B4 The number of pairs is2n+1. The degree conditionforcesP to have degreen and leading coefficient±1;we may count pairs in whichP has leading coefficient1 as long as we multiply by2 afterward.

Factor both sides:

(P (X) + Q(X)i)(P (X) − Q(X)i)

=n−1∏

j=0

(X − exp(2πi(2j + 1)/(4n)))

·n−1∏

j=0

(X + exp(2πi(2j + 1)/(4n))).

Then each choice ofP, Q corresponds to equatingP (X) + Q(X)i with the product of somen factors onthe right, in which we choose exactly of the two factorsfor eachj = 0, . . . , n−1. (We must take exactlyn fac-tors because as a polynomial inX with complex coeffi-cients,P (X) + Q(X)i has degree exactlyn. We mustchoose one for eachj to ensure thatP (X)+Q(X)i andP (X) − Q(X)i are complex conjugates, so thatP, Qhave real coefficients.) Thus there are2n such pairs;multiplying by 2 to allowP to have leading coefficient−1 yields the desired result.

Remark: If we allow P andQ to have complex co-efficients but still requiredeg(P ) > deg(Q), then thenumber of pairs increases to2

(

2nn

)

, as we may chooseany n of the 2n factors ofX2n + 1 to use to formP (X) + Q(X)i.

B5 Forn an integer, we have⌊

nk

⌋

= n−jk for j the unique

integer in 0, . . . , k − 1 congruent ton modulo k;hence

k−1∏

j=0

(

⌊n

k

⌋

− n − j

k

)

= 0.

By expanding this out, we obtain the desired polynomi-alsP0(n), . . . , Pk−1(n).


5

Remark: Variants of this solution are possible that con-struct thePi less explicitly, using Lagrange interpola-tion or Vandermonde determinants.

B6 (Suggested by Oleg Golberg) Assumen ≥ 2, or elsethe problem is trivially false. Throughout this proof,anyCi will be a positive constant whose exact value isimmaterial. As in the proof of Stirling’s approximation,we estimate for any fixedc ∈ R,

n∑

i=1

(i + c) log i =1

2n2 log n − 1

4n2 + O(n log n)

by comparing the sum to an integral. This gives

nn2/2−C1ne−n2/4 ≤ 11+c22+c · · ·nn+c

≤ nn2/2+C2ne−n2/4.

We now interpretf(n) as counting the number ofn-tuples(a1, . . . , an) of nonnegative integers such that

a11! + · · · + ann! = n!.

For an upper bound onf(n), we use the inequalities0 ≤ ai ≤ n!/i! to deduce that there are at mostn!/i! +1 ≤ 2(n!/i!) choices forai. Hence

f(n) ≤ 2n n!

1!· · · n!

n!

= 2n2132 · · ·nn−1

≤ nn2/2+C3ne−n2/4.

For a lower bound onf(n), we note that if0 ≤ ai <(n − 1)!/i! for i = 2, . . . , n − 1 andan = 0, then0 ≤a22!+ · · ·+ann! ≤ n!, so there is a unique choice ofa1

to complete this to a solution ofa11!+ · · ·+ann! = n!.Hence

f(n) ≥ (n − 1)!

2!· · · (n − 1)!

(n − 1)!

= 3142 · · · (n − 1)n−3

≥ nn2/2+C4ne−n2/4.




A1 The function g(x) = f(x, 0) works. Substituting(x, y, z) = (0, 0, 0) into the given functional equationyields f(0, 0) = 0, whence substituting(x, y, z) =(x, 0, 0) yieldsf(x, 0) + f(0, x) = 0. Finally, substi-tuting(x, y, z) = (x, y, 0) yieldsf(x, y) = −f(y, 0)−f(0, x) = g(x) − g(y).

Remark: A similar argument shows that the possiblefunctionsg are precisely those of the formf(x, 0) + cfor somec.

A2 Barbara wins using one of the following strategies.

First solution: Pair each entry of the first row with theentry directly below it in the second row. If Alan everwrites a number in one of the first two rows, Barbarawrites the same number in the other entry in the pair. IfAlan writes a number anywhere other than the first tworows, Barbara does likewise. At the end, the resultingmatrix will have two identical rows, so its determinantwill be zero.

Second solution: (by Manjul Bhargava) WheneverAlan writes a numberx in an entry in some row, Bar-bara writes−x in some other entry in the same row. Atthe end, the resulting matrix will have all rows summingto zero, so it cannot have full rank.

A3 We first prove that the process stops. Note first thatthe producta1 · · · an remains constant, becauseajak =gcd(aj , ak) lcm(aj , ak). Moreover, the last number inthe sequence can never decrease, because it is alwaysreplaced by its least common multiple with anothernumber. Since it is bounded above (by the product ofall of the numbers), the last number must eventuallyreach its maximum value, after which it remains con-stant throughout. After this happens, the next-to-lastnumber will never decrease, so it eventually becomesconstant, and so on. After finitely many steps, all of thenumbers will achieve their final values, so no more stepswill be possible. This only happens whenaj dividesak

for all pairsj < k.

We next check that there is only one possible finalsequence. Forp a prime andm a nonnegative in-teger, we claim that the number of integers in thelist divisible by pm never changes. To see this, sup-pose we replaceaj , ak by gcd(aj , ak), lcm(aj , ak). Ifneither of aj , ak is divisible by pm, then neither ofgcd(aj , ak), lcm(aj , ak) is either. If exactly oneaj , ak

is divisible bypm, thenlcm(aj , ak) is divisible bypm

butgcd(aj , ak) is not.

gcd(aj , ak), lcm(aj , ak) are as well.

If we started out with exactlyh numbers not divisibleby pm, then in the final sequencea′

1, . . . , a′n, the num-

bersa′h+1, . . . , a

′n are divisible bypm while the num-

bersa′1, . . . , a

′h are not. Repeating this argument for

each pair(p, m) such thatpm divides the initial producta1, . . . , an, we can determine the exact prime factoriza-tion of each ofa′

1, . . . , a′n. This proves that the final

sequence is unique.

Remark: (by David Savitt and Noam Elkies) Here aretwo other ways to prove the termination. One is to ob-serve that

∏

j ajj is strictly increasing at each step, and

bounded above by(a1 · · · an)n. The other is to noticethat a1 is nonincreasing but always positive, so even-tually becomes constant; thena2 is nonincreasing butalways positive, and so on.

Reinterpretation: For eachp, consider the sequenceconsisting of the exponents ofp in the prime factoriza-tions ofa1, . . . , an. At each step, we pick two positionsi andj such that the exponents of some primep are inthe wrong order at positionsi andj. We then sort thesetwo position into the correct order for every primep si-multaneously.

It is clear that this can only terminate with all se-quences being sorted into the correct order. We muststill check that the process terminates; however, sinceall but finitely many of the exponent sequences consistof all zeroes, and each step makes a nontrivial switchin at least one of the other exponent sequences, it isenough to check the case of a single exponent sequence.This can be done as in the first solution.

Remark: Abhinav Kumar suggests the following proofthat the process always terminates in at most

(

n2

)

steps.(This is a variant of the worst-case analysis of thebub-ble sortalgorithm.)

Consider the number of pairs(k, l) with 1 ≤ k < l ≤ nsuch thatak does not divideal (call thesebad pairs). Ateach step, we find one bad pair(i, j) and eliminate it,and we do not touch any pairs that do not involve eitheri or j. If i < k < j, then neither of the pairs(i, k)and(k, j) can become bad, becauseai is replaced by adivisor of itself, whileaj is replaced by a multiple ofitself. If k < i, then(k, i) can only become a bad pairif ak dividedai but notaj , in which case(k, j) stopsbeing bad. Similarly, ifk > j, then(i, k) and (j, k)either stay the same or switch status. Hence the numberof bad pairs goes down by at least 1 each time; since itis at most

(

n2

)

to begin with, this is an upper bound forthe number of steps.

Remark: This problem is closely related to the classi-fication theorem for finite abelian groups. Namely, ifa1, . . . , an anda′

1, . . . , a′n are the sequences obtained


at two different steps in the process, then the abeliangroupsZ/a1Z×· · ·×Z/anZ andZ/a′

1Z×· · ·×Z/a′nZ

are isomorphic. The final sequence gives a canonicalpresentation of this group; the terms of this sequenceare called theelementary divisorsor invariant factorsof the group.

Remark: (by Tom Belulovich) Alattice is a partiallyordered setL in which for any twox, y ∈ L, there is aunique minimal elementz with z ≥ x andz ≥ y, calledthe join and denotedx ∧ y, and there is a unique max-imal elementz with z ≤ x andz ≤ y, called themeetand denotedx∨ y. In terms of a latticeL, one can posethe following generalization of the given problem. Startwith a1, . . . , an ∈ L. If i < j but ai 6≤ aj , it is per-mitted to replaceai, aj by ai ∨aj , ai ∧aj , respectively.The same argument as above shows that this always ter-minates in at most

(

n2

)

steps. The question is, underwhat conditions on the latticeL is the final sequenceuniquely determined by the initial sequence?

It turns out that this holds if and only ifL is distributive,i.e., for anyx, y, z ∈ L,

x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).

(This is equivalent to the same axiom with the oper-ations interchanged.) For example, ifL is a Booleanalgebra, i.e., the set of subsets of a given setS underinclusion, then∧ is union,∨ is intersection, and the dis-tributive law holds. Conversely, any finite distributivelattice is contained in a Boolean algebra by a theoremof Birkhoff. The correspondence takes eachx ∈ L tothe set ofy ∈ L such thatx ≥ y andy cannot be writtenas a join of two elements ofL \ y. (See for instanceBirkhoff, Lattice Theory, Amer. Math. Soc., 1967.)

On one hand, ifL is distributive, it can be shown thatthej-th term of the final sequence is equal to the meet ofai1∧· · ·∧aij

over all sequences1 ≤ i1 < · · · < ij ≤ n.For instance, this can be checked by forming the small-est subsetL′ of L containinga1, . . . , an and closed un-der meet and join, then embeddingL′ into a Booleanalgebra using Birkhoff’s theorem, then checking theclaim for all Boolean algebras. It can also be checkeddirectly (as suggested by Nghi Nguyen) by showing thatfor j = 1, . . . , n, the meet of all joins ofj-element sub-sets ofa1, . . . , an is invariant at each step.

On the other hand, a lattice fails to be distributive ifand only if it contains five elementsa, b, c, 0, 1 such thateither the only relations among them are implied by

1 ≥ a, b, c ≥ 0

(this lattice is sometimes called thediamond), or theonly relations among them are implied by

1 ≥ a ≥ b ≥ 0, 1 ≥ c ≥ 0

(this lattice is sometimes called thepentagon). (For aproof, see the Birkhoff reference given above.) For each

of these examples, the initial sequencea, b, c fails todetermine the final sequence; for the diamond, we canend up with0, ∗, 1 for any of∗ = a, b, c, whereas for thepentagon we can end up with0, ∗, 1 for any of∗ = a, b.

Consequently, the final sequence is determined by theinitial sequence if and only ifL is distributive.

A4 The sum diverges. From the definition,f(x) = x on[1, e], x ln x on (e, ee], x ln x ln lnx on (ee, eee

], and soforth. It follows that on[1,∞), f is positive, continu-ous, and increasing. Thus

∑∞n=1

1f(n) , if it converges,

is bounded below by∫ ∞1

dxf(x) ; it suffices to prove that

the integral diverges.

Write ln1 x = lnx and lnk x = ln(lnk−1 x) fork ≥ 2; similarly write exp1 x = ex and expk x =

eexpk−1 x. If we write y = lnk x, then x =expk y anddx = (expk y)(expk−1 y) · · · (exp1 y)dy =

x(ln1 x) · · · (lnk−1 x)dy. Now on [expk−1 1, expk 1],we havef(x) = x(ln1 x) · · · (lnk−1 x), and thus sub-stitutingy = lnk x yields

∫ expk 1

expk−1 1

dx

f(x)=

∫ 1

0

dy = 1.

It follows that∫ ∞1

dxf(x) =

∑∞k=1

∫ expk 1

expk−1 1dx

f(x) di-verges, as desired.

A5 Form the polynomialP (z) = f(z) + ig(z) with com-plex coefficients. It suffices to prove thatP has degreeat leastn − 1, as then one off, g must have degree atleastn − 1.

By replacingP (z) with aP (z) + b for suitablea, b ∈C, we can force the regularn-gon to have verticesζn, ζ2

n, . . . , ζnn for ζn = exp(2πi/n). It thus suffices to

check that there cannot exist a polynomialP (z) of de-gree at mostn−2 such thatP (i) = ζi

n for i = 1, . . . , n.

We will prove more generally that for any complexnumbert /∈ 0, 1, and any integerm ≥ 1, any poly-nomialQ(z) for whichQ(i) = ti for i = 1, . . . , m hasdegree at leastm−1. There are several ways to do this.

First solution: If Q(z) has degreed and leading coef-ficient c, thenR(z) = Q(z + 1) − tQ(z) has degreedand leading coefficient(1 − t)c. However, by hypoth-esis,R(z) has the distinct roots1, 2, . . . , m − 1, so wemust haved ≥ m − 1.

Second solution: We proceed by induction onm. Forthe base casem = 1, we haveQ(1) = t1 6= 0, soQmust be nonzero, and so its degree is at least0. Giventhe assertion form − 1, if Q(i) = ti for i = 1, . . . , m,then the polynomialR(z) = (t−1)−1(Q(z+1)−Q(z))has degree one less than that ofQ, and satisfiesR(i) =ti for i = 1, . . . , m − 1. SinceR must have degree atleastm − 2 by the induction hypothesis,Q must havedegree at leastm − 1.

Third solution: We use the method offinite differ-ences(as in the second solution) but without induction.


Namely, the(m− 1)-st finite difference ofP evaluatedat 1 equals

m−1∑

j=0

(−1)j

(

m − 1

j

)

Q(m − j) = t(1 − t)m−1 6= 0,

which is impossible ifQ has degree less thanm − 1.

Remark: One can also establish the claim by comput-ing a Vandermonde-type determinant, or by using theLagrange interpolation formula to compute the leadingcoefficient ofQ.

A6 For notational convenience, we will interpret the prob-lem as allowing the empty subsequence, whose productis the identity element of the group. To solve the prob-lem in the interpretation where the empty subsequenceis not allowed, simply append the identity element tothe sequence given by one of the following solutions.

First solution: Put n = |G|. We will say that a se-quenceS producesan elementg ∈ G if g occurs as theproduct of some subsequence ofS. Let H be the set ofelements produced by the sequenceS.

Start withS equal to the empty sequence. If at any pointthe setH−1H = h1h2 : h−1

1 , h2 ∈ H fails to be allof G, extendS by appending an elementg of G notin H−1H . ThenHg ∩ H must be empty, otherwisethere would be an equation of the formh1g = h2 withh1, h2 ∈ G, or g = h−1

1 h2, a contradiction. Thus wecan extendS by one element and double the size ofH .

After k ≤ log2 n steps, we must obtain a sequenceS = a1, . . . , ak for which H−1H = G. Then the se-quencea−1

k , . . . , a−11 , a1, . . . , ak produces all ofG and

has length at most(2/ ln 2) lnn.

Second solution:

Put m = |H |. We will show that we can append oneelementg to S so that the resulting sequence ofk + 1elements will produce at least2m − m2/n elements ofG. To see this, we compute

∑

g∈G

|H ∪ Hg| =∑

g∈G

(|H | + |Hg| − |H ∩ Hg|)

= 2mn −∑

g∈G

|H ∩ Hg|

= 2mn − |(g, h) ∈ G2 : h ∈ H ∩ Hg|= 2mn −

∑

h∈H

|g ∈ G : h ∈ Hg|

= 2mn −∑

h∈H

|H−1h|

= 2mn − m2.

By the pigeonhole principle, we have|H∪Hg| ≥ 2m−m2/n for some choice ofg, as claimed.

In other words, by extending the sequence by one el-ement, we can replace the ratios = 1 − m/n (i.e.,

the fraction of elements ofG not generated byS) by aquantity no greater than

1 − (2m − m2/n)/n = s2.

We start out withk = 0 ands = 1− 1/n; afterk steps,we haves ≤ (1− 1/n)2

k

. It is enough to prove that forsomec > 0, we can always find an integerk ≤ c lnnsuch that

(

1 − 1

n

)2k

<1

n,

as then we haven − m < 1 and henceH = G.

To obtain this last inequality, put

k = ⌊2 log2 n⌋ < (2/ ln 2) lnn,

so that2k+1 ≥ n2. From the facts thatlnn ≤ ln 2 +(n − 2)/2 ≤ n/2 and ln(1 − 1/n) < −1/n for alln ≥ 2, we have

2k ln

(

1 − 1

n

)

< −n2

2n= −n

2< − lnn,

yielding the desired inequality.

Remark: An alternate approach in the second solutionis to distinguish betwen the cases ofH small (i.e.,m <n1/2, in which casem can be replaced by a value noless than2m − 1) andH large. This strategy is used ina number of recent results of Bourgain, Tao, Helfgott,and others onsmall doublingorsmall triplingof subsetsof finite groups.

In the second solution, if we avoid the rather weak in-equalitylnn ≤ n/2, we instead get sequences of lengthlog2(n lnn) = log2(n) + log2(lnn). This is close tooptimal: one cannot use fewer thanlog2 n terms be-cause the number of subsequences must be at leastn.

B1 There are at most two such points. For example,the points(0, 0) and (1, 0) lie on a circle with center(1/2, x) for any real numberx, not necessarily rational.

On the other hand, supposeP = (a, b), Q =(c, d), R = (e, f) are three rational points that lie ona circle. The midpointM of the sidePQ is ((a +c)/2, (b+ d)/2), which is again rational. Moreover, theslope of the linePQ is (d−b)/(c−a), so the slope of theline throughM perpendicular toPQ is (a− c)/(b−d),which is rational or infinite.

Similarly, if N is the midpoint ofQR, thenN is a ratio-nal point and the line throughN perpendicular toQRhas rational slope. The center of the circle lies on bothof these lines, so its coordinates(g, h) satisfy two linearequations with rational coefficients, sayAg + Bh = CandDg + Eh = F . Moreover, these equations have aunique solution. That solution must then be

g = (CE − BD)/(AE − BD)

h = (AF − BC)/(AE − BD)


(by elementary algebra, or Cramer’s rule), so the centerof the circle is rational. This proves the desired result.

Remark: The above solution is deliberately more ver-bose than is really necessary. A shorter way to say thisis that any two distinct rational points determine ara-tional line (a line of the formax + by + c = 0 witha, b, c rational), while any two nonparallel rational linesintersect at a rational point. A similar statement holdswith the rational numbers replaced by any field.

Remark: A more explicit argument is to show thatthe equation of the circle through the rational points(x1, y1), (x2, y2), (x3, y3) is

0 = det

x21 + y2

1 x1 y1 1x2

2 + y22 x2 y2 1

x23 + y2

3 x3 y3 1x2 + y2 x y 1

which has the forma(x2 + y2) + dx + ey + f =0 for a, d, e, f rational. The center of this circle is(−d/(2a),−e/(2a)), which is again a rational point.

B2 We claim thatFn(x) = (lnx − an)xn/n!, wherean =∑n

k=1 1/k. Indeed, temporarily writeGn(x) = (lnx−an)xn/n! for x > 0 andn ≥ 1; thenlimx→0 Gn(x) =0 andG′

n(x) = (lnx − an + 1/n)xn−1/(n − 1)! =Gn−1(x), and the claim follows by the FundamentalTheorem of Calculus and induction onn.

Given the claim, we haveFn(1) = −an/n! and so weneed to evaluate− limn→∞

an

ln n . But since the function1/x is strictly decreasing forx positive,

∑nk=2 1/k =

an − 1 is bounded below by∫ n

2 dx/x = lnn −ln 2 and above by

∫ n

1 dx/x = lnn. It follows thatlimn→∞

an

ln n = 1, and the desired limit is−1.

B3 The largest possible radius is√

22 . It will be convenient

to solve the problem for a hypercube of side length 2instead, in which case we are trying to show that thelargest radius is

√2.

Choose coordinates so that the interior of the hypercubeis the setH = [−1, 1]4 in R4. LetC be a circle centeredat the pointP . ThenC is contained both inH andits reflection acrossP ; these intersect in a rectangularparalellepiped each of whose pairs of opposite faces areat most 2 unit apart. Consequently, if we translateC sothat its center moves to the pointO = (0, 0, 0, 0) at thecenter ofH , then it remains entirely insideH .

This means that the answer we seek equals the largestpossible radius of a circleC contained inH and cen-tered at O. Let v1 = (v11, . . . , v14) and v2 =(v21, . . . , v24) be two points onC lying on radii per-pendicular to each other. Then the points of the circlecan be expressed asv1 cos θ + v2 sin θ for 0 ≤ θ < 2π.ThenC lies inH if and only if for eachi, we have

|v1i cos θ + v2i sin θ| ≤ 1 (0 ≤ θ < 2π).

In geometric terms, the vector(v1i, v2i) in R2 hasdot product at most 1 with every unit vector. Sincethis holds for the unit vector in the same direction as(v1i, v2i), we must have

v21i + v2

2i ≤ 1 (i = 1, . . . , 4).

Conversely, if this holds, then the Cauchy-Schwarz in-equality and the above analysis imply thatC lies inH .

If r is the radius ofC, then

2r2 =

4∑

i=1

v21i +

4∑

i=1

v22i

=

4∑

i=1

(v21i + v2

2i)

≤ 4,

sor ≤√

2. Since this is achieved by the circle through(1, 1, 0, 0) and(0, 0, 1, 1), it is the desired maximum.

Remark: One may similarly ask for the radius of thelargestk-dimensional ball inside ann-dimensional unithypercube; the given problem is the case(n, k) =(4, 2). Daniel Kane gives the following argument toshow that the maximum radius in this case is1

2

√

nk .

(Thanks for Noam Elkies for passing this along.)

We again scale up by a factor of 2, so that we are tryingto show that the maximum radiusr of a k-dimensionalball contained in the hypercube[−1, 1]n is

√

nk . Again,

there is no loss of generality in centering the ball at theorigin. LetT : R

k → Rn be a similitude carrying the

unit ball to this embeddedk-ball. Then there exists avectorvi ∈ R

k such that fore1, . . . , en the standard ba-sis ofRn, x·vi = T (x)·ei for all x ∈ Rk. The conditionof the problem is equivalent to requiring|vi| ≤ 1 for alli, while the radiusr of the embedded ball is determinedby the fact that for allx ∈ Rk,

r2(x · x) = T (x) · T (x) =

n∑

i=1

x · vi.

Let M be the matrix with columnsv1, . . . , vk; thenMMT = r2Ik, for Ik the k × k identity matrix. Wethen have

kr2 = Trace(r2Ik) = Trace(MMT )

= Trace(MT M) =

n∑

i=1

|vi|2

≤ n,

yielding the upper boundr ≤√

nk .

To show that this bound is optimal, it is enough to showthat one can find an orthogonal projection ofR

n ontoRk so that the projections of theei all have the samenorm (one can then rescale to get the desired configura-tion of v1, . . . , vn). We construct such a configuration


by a “smoothing” argument. Startw with any projec-tion. Let w1, . . . , wn be the projections ofe1, . . . , en.If the desired condition is not achieved, we can choosei, j such that

|wi|2 <1

n(|w1|2 + · · · + |wn|2) < |wj |2.

By precomposing with a suitable rotation that fixeseh

for h 6= i, j, we can vary|wi|, |wj | without varying|wi|2 + |wj |2 or |wh| for h 6= i, j. We can thus choosesuch a rotation to force one of|wi|2, |wj |2 to becomeequal to 1

n (|w1|2 + · · · + |wn|2). Repeating at mostn − 1 times gives the desired configuration.

B4 We use the identity given by Taylor’s theorem:

h(x + y) =

deg(h)∑

i=0

h(i)(x)

i!yi.

In this expression,h(i)(x)/i! is a polynomial inx withinteger coefficients, so its value at an integerx is aninteger.

Forx = 0, . . . , p − 1, we deduce that

h(x + p) ≡ h(x) + ph′(x) (mod p2).

(This can also be deduced more directly using the bino-mial theorem.) Since we assumedh(x) andh(x + p)are distinct modulop2, we conclude thath′(x) 6≡ 0(mod p). Sinceh′ is a polynomial with integer coeffi-cients, we haveh′(x) ≡ h′(x + mp) (mod p) for anyintegerm, and soh′(x) 6≡ 0 (mod p) for all integersx.

Now for x = 0, . . . , p2 − 1 andy = 0, . . . , p − 1, wewrite

h(x + yp2) ≡ h(x) + p2yh′(x) (mod p3).

Thush(x), h(x + p2), . . . , h(x + (p − 1)p2) run overall of the residue classes modulop3 congruent toh(x) modulo p2. Since theh(x) themselves coverall the residue classes modulop2, this proves thath(0), . . . , h(p3 − 1) are distinct modulop3.

Remark: More generally, the same proof shows thatfor any integersd, e > 1, h permutes the residue classesmodulopd if and only if it permutes the residue classesmodulope. The argument used in the proof is relatedto a general result in number theory known asHensel’slemma.

B5 The functionsf(x) = x+n andf(x) = −x+n for anyintegern clearly satisfy the condition of the problem;we claim that these are the only possiblef .

Let q = a/b be any rational number withgcd(a, b) = 1andb > 0. Forn any positive integer, we have

f(an+1bn ) − f(a

b )1bn

= bnf

(

an + 1

bn

)

− nbf(a

b

)

is an integer by the property off . Sincef is differen-tiable ata/b, the left hand side has a limit. It followsthat for sufficiently largen, both sides must be equal tosome integerc = f ′(a

b ): f(an+1bn ) = f(a

b ) + cbn . Now

c cannot be0, since otherwisef(an+1bn ) = f(a

b ) forsufficiently largen has denominatorb rather thanbn.Similarly, |c| cannot be greater than1: otherwise if wetaken = k|c| for k a sufficiently large positive integer,thenf(a

b ) + cbn has denominatorbk, contradicting the

fact thatf(an+1bn ) has denominatorbn. It follows that

c = f ′(ab ) = ±1.

Thus the derivative off at any rational number is±1.Sincef is continuously differentiable, we conclude thatf ′(x) = 1 for all realx or f ′(x) = −1 for all realx.Sincef(0) must be an integer (a rational number withdenominator1), f(x) = x + n or f(x) = −x + n forsome integern.

Remark: After showing thatf ′(q) is an integer foreachq, one can instead argue thatf ′ is a continuousfunction from the rationals to the integers, so must beconstant. One can then writef(x) = ax + b and checkthatb ∈ Z by evaluation ata = 0, and thata = ±1 byevaluation atx = 1/a.

B6 In all solutions, letFn,k be the number ofk-limited per-mutations of1, . . . , n.

First solution: (by Jacob Tsimerman) Note that anypermutation isk-limited if and only if its inverse isk-limited. Consequently, the number ofk-limited per-mutations of1, . . . , n is the same as the number ofk-limited involutions (permutations equal to their in-verses) of1, . . . , n.

We use the following fact several times: the numberof involutions of 1, . . . , n is odd if n = 0, 1 andeven otherwise. This follows from the fact that non-involutions come in pairs, so the number of involu-tions has the same parity as the number of permutations,namelyn!.

For n ≤ k + 1, all involutions arek-limited. By theprevious paragraph,Fn,k is odd forn = 0, 1 and evenfor n = 2, . . . , k + 1.

For n > k + 1, group thek-limited involutions intoclasses based on their actions onk + 2, . . . , n. Notethat for C a class andσ ∈ C, the set of elements ofA = 1, . . . , k + 1 which map intoA underσ de-pends only onC, not onσ. Call this setS(C); then thesize ofC is exactly the number of involutions ofS(C).Consequently,|C| is even unlessS(C) has at most oneelement. However, the element 1 cannot map out ofAbecause we are looking atk-limited involutions. Henceif S(C) has one element andσ ∈ C, we must haveσ(1) = 1. Sinceσ is k-limited andσ(2) cannot belongto A, we must haveσ(2) = k + 2. By induction, fori = 3, . . . , k + 1, we must haveσ(i) = k + i.

If n < 2k + 1, this shows that no classC of odd cardi-nality can exist, soFn,k must be even. Ifn ≥ 2k + 1,


the classes of odd cardinality are in bijection withk-limited involutions of2k + 2, . . . , n, soFn,k has thesame parity asFn−2k−1,k. By induction onn, we de-duce the desired result.

Second solution: (by Yufei Zhao) LetMn,k be then×n matrix with

(Mn,k)ij =

1 |i − j| ≤ k

0 otherwise.

Write det(Mn,k) as the sum over permutationsσ of1, . . . , n of (Mn,k)1σ(1) · · · (Mn,k)nσ(n) times thesignature ofσ. Thenσ contributes±1 to det(Mn,k)if σ is k-limited and 0 otherwise. We conclude that

det(Mn,k) ≡ Fn,k (mod 2).

For the rest of the solution, we interpretMn,k as a ma-trix over the field of two elements. We compute its de-terminant using linear algebra modulo 2.

We first show that forn ≥ 2k + 1,

Fn,k ≡ Fn−2k−1,k (mod 2),

provided that we interpretF0,k = 1. We do this by com-puting det(Mn,k) using row and column operations.We will verbally describe these operations for generalk, while illustrating with the examplek = 3.

To begin with,Mn,k has the following form.

1 1 1 1 0 0 0 ∅1 1 1 1 1 0 0 ∅1 1 1 1 1 1 0 ∅1 1 1 1 1 1 1 ∅0 1 1 1 1 1 1 ?0 0 1 1 1 1 1 ?0 0 0 1 1 1 1 ?

∅ ∅ ∅ ∅ ? ? ? ∗

In this presentation, the first2k + 1 rows and columnsare shown explicitly; the remaining rows and columnsare shown in a compressed format. The symbol∅ in-dicates that the unseen entries are all zeroes, while thesymbol? indicates that they are not. The symbol∗ inthe lower right corner represents the matrixFn−2k−1,k.We will preserve the unseen structure of the matrix byonly adding the firstk + 1 rows or columns to any ofthe others.

We first add row 1 to each of rows2, . . . , k + 1.

1 1 1 1 0 0 0 ∅0 0 0 0 1 0 0 ∅0 0 0 0 1 1 0 ∅0 0 0 0 1 1 1 ∅0 1 1 1 1 1 1 ?

0 0 1 1 1 1 1 ?

0 0 0 1 1 1 1 ?

∅ ∅ ∅ ∅ ? ? ? ∗

We next add column 1 to each of columns2, . . . , k + 1.

1 0 0 0 0 0 0 ∅0 0 0 0 1 0 0 ∅0 0 0 0 1 1 0 ∅0 0 0 0 1 1 1 ∅0 1 1 1 1 1 1 ?

0 0 1 1 1 1 1 ?

0 0 0 1 1 1 1 ?

∅ ∅ ∅ ∅ ? ? ? ∗

For i = 2, for each ofj = i + 1, . . . , 2k + 1 for whichthe(j, k + i)-entry is nonzero, add rowi to row j.

1 0 0 0 0 0 0 ∅0 0 0 0 1 0 0 ∅0 0 0 0 0 1 0 ∅0 0 0 0 0 1 1 ∅0 1 1 1 0 1 1 ?

0 0 1 1 0 1 1 ?

0 0 0 1 0 1 1 ?

∅ ∅ ∅ ∅ ∅ ? ? ∗

Repeat the previous step fori = 3, . . . , k +1 in succes-sion.

1 0 0 0 0 0 0 ∅0 0 0 0 1 0 0 ∅0 0 0 0 0 1 0 ∅0 0 0 0 0 0 1 ∅0 1 1 1 0 0 0 ?

0 0 1 1 0 0 0 ?

0 0 0 1 0 0 0 ?

∅ ∅ ∅ ∅ ∅ ∅ ∅ ∗

Repeat the two previous steps with the roles of the rowsand columns reversed. That is, fori = 2, . . . , k + 1, foreach ofj = i + 1, . . . , 2k + 1 for which the(j, k + i)-entry is nonzero, add rowi to row j.

1 0 0 0 0 0 0 ∅0 0 0 0 1 0 0 ∅0 0 0 0 0 1 0 ∅0 0 0 0 0 0 1 ∅0 1 0 0 0 0 0 ∅0 0 1 0 0 0 0 ∅0 0 0 1 0 0 0 ∅∅ ∅ ∅ ∅ ∅ ∅ ∅ ∗

We now have a block diagonal matrix in which the topleft block is a(2k + 1)× (2k + 1) matrix with nonzerodeterminant (it results from reordering the rows of theidentity matrix), the bottom right block isMn−2k−1,k,and the other two blocks are zero. We conclude that

det(Mn,k) ≡ det(Mn−2k−1,k) (mod 2),


proving the desired congruence.

To prove the desired result, we must now check thatF0,k, F1,k are odd andF2,k, . . . , F2k,k are even. Forn = 0, . . . , k + 1, the matrixMn,k consists of all ones,so its determinant is 1 ifn = 0, 1 and 0 otherwise. (Al-ternatively, we haveFn,k = n! for n = 0, . . . , k + 1,since every permutation of1, . . . , n is k-limited.) Forn = k + 2, . . . , 2k, observe that rowsk andk + 1 ofMn,k both consist of all ones, sodet(Mn,k) = 0 asdesired.

Third solution: (by Tom Belulovich) DefineMn,k asin the second solution. We provedet(Mn,k) is odd forn ≡ 0, 1 (mod 2k+1) and even otherwise, by directlydetermining whether or notMn,k is invertible as a ma-trix over the field of two elements.

Let ri denote rowi of Mn,k. We first check that ifn ≡2, . . . , 2k (mod 2k + 1), thenMn,k is not invertible.In this case, we can find integers0 ≤ a < b ≤ k suchthatn + a + b ≡ 0 (mod 2k + 1). Putj = (n + a +b)/(2k + 1). We can then write the all-ones vector bothas

j−1∑

i=0

rk+1−a+(2k+1)i

and as

j−1∑

i=0

rk+1−b+(2k+1)i.

HenceMn,k is not invertible.

We next check that ifn ≡ 0, 1 (mod 2k + 1), thenMn,k is invertible. Suppose thata1, . . . , an are scalars

such thata1r1 + · · · + anrn is the zero vector. Them-th coordinate of this vector equalsam−k + · · ·+ am+k,where we regardai as zero ifi /∈ 1, . . . , n. By com-paring consecutive coordinates, we obtain

am−k = am+k+1 (1 ≤ m < n).

In particular, theai repeat with period2k + 1. Takingm = 1, . . . , k further yields that

ak+2 = · · · = a2k+1 = 0

while takingm = n − k, . . . , n − 1 yields

an−2k = · · · = an−1−k = 0.

Forn ≡ 0 (mod 2k + 1), the latter can be rewritten as

a1 = · · · = ak = 0

whereas forn ≡ 1 (mod 2k+1), it can be rewritten as

a2 = · · · = ak+1 = 0.

In either case, since we also have

a1 + · · · + a2k+1 = 0from the(k+1)-st coordinate, we deduce that all of theai must be zero, and soMn,k must be invertible.

Remark: The matricesMn,k are examples ofbandedmatrices, which occur frequently in numerical appli-cations of linear algebra. They are also examples ofToeplitz matrices.


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